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A"},"content":{"rendered":""},"excerpt":{"rendered":""},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":60,"niveau_difficulte":"Intermediate","categorie":"Full Practice Exam 1","nombre_questions":30,"score_minimum":80,"ordre_questions":"Fixe","questions":[{"id":"cmllmqqxz001sju04di97x8qt","texte_question":"

What is the instantaneous rate of change for the function $$g(x)=\\frac{x^{3}-3x^{2}+x-2}{x-1},$$ when $$x=2$$?

","type_question":"QCM","explication":"$17","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-5$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-1$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$5$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqxz001xju047p84logl","texte_question":"$18","type_question":"QCM","explication":"

The total number of bicycles is the area under the rate curve from $$t=0$$ to $$t=15$$ hours.

From the graph, the rate is approximately $$30$$ bikes per minute on average up to $$15$$ hours.

Convert $$15$$ hours to minutes:

$$15 \\cdot 60 = 900$$ minutes.

Now multiply average rate by total minutes:

$$30 \\cdot 900 = 27000$$

Final answer: $$27000$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$45$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$720$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$1080$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$27000$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqxz0022ju041olmk0za","texte_question":"

Evaluate the definite integral:

$$\\int_{2}^{5}\\frac{3x}{x^4}dx$$

","type_question":"QCM","explication":"$19","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{63}{200}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{9}{20}$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{27}{40}$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{3}{4}$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqxz0027ju044awot15w","texte_question":"$1a","type_question":"QCM","explication":"

Compute the average rate of change of $$g$$ on $$[1,3]:$$

$$\\frac{g(3)-g(1)}{3-1}=\\frac{15-4}{2}=\\frac{11}{2}$$

Since $$\\frac{11}{2}>0,$$ the average rate of change is positive.

Check the other options:

$$A)$$ Not guaranteed. A continuous function can have inflection points between the given data values.

$$C)$$ Not guaranteed. The derivative at $$x=2$$ depends on the local behavior near $$2,$$ not just table values.

$$D)$$ Not guaranteed. The slopes between consecutive points are

$$\\frac{g(2)-g(1)}{2-1}=\\frac{7-4}{1}=3$$

and

$$\\frac{g(3)-g(2)}{3-2}=\\frac{15-7}{1}=8.$$

Since $$3\\neq 8,$$ the data do not follow a single line, and $$g$$ need not be linear.

Therefore, the only statement that must be true is $$B.$$

Final answer: The average rate of change of $$g$$ on $$[1,3]$$ is positive

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$g$$ has no inflection points on $$[1,3]$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"The average rate of change of $$g$$ on $$[1,3]$$ is positive","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$g'(2)=4$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$g$$ is linear on $$[1,3]$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy0002cju04zaw4rejo","texte_question":"

Evaluate the definite integral:

$$\\int_{\\frac{\\pi}{4}}^{\\frac{\\pi}{2}}\\frac{\\sec x\\tan x}{\\cos x}dx$$

","type_question":"QCM","explication":"$1b","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$e-1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$1-e$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$e$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"This integral does not have a finite value.","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy0002hju04gyzkhys6","texte_question":"

At the point $$(2,1)$$, what is the slope of the curve defined implicitly by:

$$x^2+4y^2+2xy-6x-8y=0?$$

","type_question":"QCM","explication":"$1c","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-\\frac{3}{2}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-\\frac{1}{2}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$0$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"It is undefined","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy0002mju04lpj5oit4","texte_question":"

Evaluate the integral:

$$\\int_0^1\\sqrt{x}(3x^2-2x+1)dx$$

","type_question":"QCM","explication":"$1d","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{8}{15}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{76}{105}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{24}{35}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{32}{35}$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy0002rju04glt37l9d","texte_question":"

The radius of a sphere is increasing at a rate of $$2$$ inches per minute. When the surface area of the sphere is $$16\\pi$$ square inches, how fast is the volume increasing in cubic inches per minute?

","type_question":"QCM","explication":"

Surface area of a sphere:

$$S=4\\pi r^2.$$

Given:

$$4\\pi r^2=16\\pi.$$

Divide both sides by $$4\\pi$$:

$$r^2=4.$$

$$r=2.$$

Volume of a sphere:

$$V=\\frac{4}{3}\\pi r^3.$$

Differentiate with respect to $$t$$:

$$\\frac{dV}{dt}=4\\pi r^2\\frac{dr}{dt}.$$

Substitute $$r=2$$ and $$\\frac{dr}{dt}=2$$:

$$\\frac{dV}{dt}=4\\pi(2)^2(2).$$

$$=4\\pi(4)(2).$$

$$=32\\pi.$$

Final answer: $$32\\pi$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$8\\pi$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$16\\pi$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$32\\pi$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$64\\pi$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy0002wju04wtq42f73","texte_question":"$1e","type_question":"QCM","explication":"

Find the intersection of $$y=x^3$$ and $$y=2x+4$$:

$$x^3=2x+4.$$

$$x^3-2x-4=0.$$

Test $$x=2$$:

$$(2)^3-2(2)-4=8-4-4=0,$$

so the curves intersect at $$x=2$$.

At this point,

$$y=2(2)+4=8.$$

From the graph, the shaded region starts at the $$y$$-axis, so the left boundary is $$x=0$$.

On the interval $$0\\le x\\le2$$, the top curve is $$y=2x+4$$ and the bottom curve is $$y=x^3$$.

Area between curves:

$$\\int_0^2\\big((2x+4)-(x^3)\\big)dx=\\int_0^2(2x+4-x^3)dx.$$

Final answer: $$\\int_0^2(2x+4-x^3)dx$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\int_0^2(2x+4-x^3)dx$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\int_2^0(2x+4-x^3)dx$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\int_0^2(x^3-2x-4)dx$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\int_0^8(2x+4-x^3)dx$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy00031ju04sne452dy","texte_question":"

What is the average rate of change of $$g(x)=x^3-2x^2+5x-1$$ over the interval $$[1,4]$$?

","type_question":"QCM","explication":"

Average rate of change on $$[1,4]$$ is:

$$\\frac{g(4)-g(1)}{4-1}.$$

Compute $$g(4)$$:

$$g(4)=(4)^3-2(4)^2+5(4)-1.$$

$$g(4)=64-2(16)+20-1.$$

$$g(4)=64-32+20-1.$$

$$g(4)=51.$$

Compute $$g(1)$$:

$$g(1)=(1)^3-2(1)^2+5(1)-1.$$

$$g(1)=1-2+5-1.$$

$$g(1)=3.$$

Now compute the average rate of change:

$$\\frac{g(4)-g(1)}{4-1}=\\frac{51-3}{3}=\\frac{48}{3}=16.$$

The value $$16$$ is not listed in the options, but it is the correct average rate of change.

Final answer: $$16$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$6$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$9$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$12$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$16$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy00036ju04996oqd1w","texte_question":"

If the graph of the second derivative of a function $$f$$ is a line with nonzero slope, what type of function could $$f$$ be?

","type_question":"QCM","explication":"

If the graph of the second derivative is a line with nonzero slope, then $$f''(x)$$ is a linear function.

A linear function has the form

$$f''(x)=ax+b$$

where $$a\\neq0.$$

Integrate once to find $$f'(x)$$:

$$f'(x)=\\int(ax+b)dx=\\frac{a}{2}x^2+bx+C.$$

This is a quadratic function.

Integrate again to find $$f(x)$$:

$$f(x)=\\int\\left(\\frac{a}{2}x^2+bx+C\\right)dx=\\frac{a}{6}x^3+\\frac{b}{2}x^2+Cx+D.$$

This is a cubic function.

Therefore, if the second derivative is linear with nonzero slope, then $$f$$ must be a cubic function.

Final answer: Cubic

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Linear","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"Quadratic","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"Cubic","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"Quartic","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy0003bju04fcd6cepe","texte_question":"

Let the function $$f$$ be defined by

$$f(x)=\\sqrt{4-x}$$ for $$x<1$$

$$f(x)=x^2$$ for $$x\\ge1$$

What is the average value of $$f$$ on the interval $$[0,3]$$?

","type_question":"QCM","explication":"$1f","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{7}{3}$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{8}{3}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{3}\\left(14-2\\sqrt{3}\\right)$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{10}{3}$$ ","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy0003gju043fon3mmh","texte_question":"$20","type_question":"QCM","explication":"

Horizontal tangent at $$x=2$$ means the slope is $$0$$, so $$g'(2)=0.$$

Concave up at $$x=2$$ means the second derivative is positive, so $$g''(2)>0.$$

From the graph, the point at $$x=2$$ lies below the $$x$$-axis, so $$g(2)<0.$$

Now compare the three values:

$$g(2)<0,$$

$$g'(2)=0,$$

$$g''(2)>0.$$

Therefore the order is:

$$g(2)<g'(2)<g''(2).$$

Final answer: $$g(2)<g'(2)<g''(2)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$g''(2)Let $$f$$ be continuous on $$[-1,5]$$ and differentiable on $$(-1,5)$$. If $$f(-1)=2$$ and $$f(5)=8$$, which statement is guaranteed by the Mean Value Theorem?

","type_question":"QCM","explication":"

Because $$f$$ is continuous on $$[-1,5]$$ and differentiable on $$(-1,5)$$, the Mean Value Theorem applies.

Compute the average rate of change on $$[-1,5]$$:

$$\\frac{f(5)-f(-1)}{5-(-1)}=\\frac{8-2}{6}=\\frac{6}{6}=1.$$

By the Mean Value Theorem, there exists at least one number $$c\\in(-1,5)$$ such that

$$f'(c)=\\frac{f(5)-f(-1)}{5-(-1)}=1.$$

So, the guaranteed statement is $$C$$.

Final answer: There exists some $$c\\in(-1,5)$$ such that $$f'(c)=1$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$f'(2)=1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$f'(x)=1$$ for all $$x\\in(-1,5)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"There exists some $$c\\in(-1,5)$$ such that $$f'(c)=1$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$f$$ is increasing on $$[-1,5]$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy1003qju045016eiln","texte_question":"

Evaluate:

$$\\frac{d}{dx}\\left(\\int_{0}^{x^2}e^tdt\\right)$$

","type_question":"QCM","explication":"

Let

$$F(x)=\\int_{0}^{x^2}e^tdt.$$

By the Fundamental Theorem of Calculus and the chain rule,

$$\\frac{d}{dx}\\int_{0}^{g(x)}e^tdt=e^{g(x)}g'(x).$$

Here

$$g(x)=x^2$$

$$g'(x)=2x.$$

Therefore,

$$\\frac{d}{dx}\\left(\\int_{0}^{x^2}e^tdt\\right)=e^{x^2}\\cdot2x.$$

Simplify:

$$2xe^{x^2}.$$

Final answer: $$2xe^{x^2}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$e^{x^2}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2xe^{x^2}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$x^2e^x$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$e^x$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy1003vju047ybearok","texte_question":"

Assume $$f(x)=e^x.$$ If the rate of change of $$f$$ at $$x=c$$ is $$e^3$$ times the rate of change at $$x=1,$$ then $$c=$$?

","type_question":"QCM","explication":"

The rate of change of $$f$$ is its derivative.

Since

$$f(x)=e^x,$$

we have

$$f'(x)=e^x.$$

The rate of change at $$x=c$$ is

$$f'(c)=e^c.$$

The rate of change at $$x=1$$ is

$$f'(1)=e^1=e.$$

We are given

$$e^c=e^3\\cdot e.$$

Simplify the right side:

$$e^3\\cdot e=e^4.$$

$$e^c=e^4.$$

Therefore

$$c=4.$$

Final answer: $$4$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$3$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$4$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$5$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy10040ju0457t6fmit","texte_question":"$21","type_question":"QCM","explication":"

Using the chain rule,

$$h'(x)=f'(g(x))\\cdot g'(x).$$

We evaluate at each given value of $$x.$$

At $$x=1:$$

$$g(1)=2,$$ so $$f'(g(1))=f'(2)=0.$$

$$g'(1)=1.$$

Thus

$$h'(1)=0\\cdot 1=0.$$

At $$x=2:$$

$$g(2)=4,$$ so $$f'(g(2))=f'(4)=3.$$

$$g'(2)=-2.$$

Thus

$$h'(2)=3\\cdot(-2)=-6.$$

At $$x=3:$$

$$g(3)=1,$$ so $$f'(g(3))=f'(1)=-1.$$

$$g'(3)=1.$$

Thus

$$h'(3)=-1\\cdot 1=-1.$$

At $$x=4:$$

$$g(4)=3,$$ so $$f'(g(4))=f'(3)=2.$$

$$g'(4)=-1.$$

Thus

$$h'(4)=2\\cdot(-1)=-2.$$

The only value where $$h'(x)=0$$ is

$$x=1.$$

Final answer: $$1$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$3$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$4$$ ","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy10045ju04l5669szp","texte_question":"

Let $$f$$ be differentiable on $$[0,6]$$ with $$f(0)=2$$ and $$f(6)=2$$. If the equation $$f(x)=5$$ has exactly two solutions in $$(0,6),$$ which statement must be true?

","type_question":"QCM","explication":"$22","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$f$$ is increasing on $$(0,6)$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$f'(x)=0$$ for some $$x\\in(0,6)$$ ","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$f$$ has an absolute minimum at $$x=3$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$f(x)\\ge 2$$ for all $$x\\in[0,6]$$ ","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy1004aju042wh6aotb","texte_question":"

Find the slope of the normal line to the curve $$y=\\sqrt{x+1}$$ at the point $$(3,2).$$

","type_question":"QCM","explication":"

First find the derivative of $$y=\\sqrt{x+1}.$$

Rewrite:

$$y=(x+1)^{\\frac{1}{2}}.$$

Differentiate:

$$y'=\\frac{1}{2}(x+1)^{-\\frac{1}{2}}.$$

Simplify:

$$y'=\\frac{1}{2\\sqrt{x+1}}.$$

Evaluate at $$x=3:$$

$$y'(3)=\\frac{1}{2\\sqrt{4}}=\\frac{1}{2\\cdot 2}=\\frac{1}{4}.$$

So the slope of the tangent line at $$(3,2)$$ is

$$\\frac{1}{4}.$$

The slope of the normal line is the negative reciprocal:

$$m_{normal}=-\\frac{1}{\\frac{1}{4}}=-4.$$

Final answer: $$-4$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-4$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-2$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-\\frac{1}{2}$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-\\frac{1}{4}$$ ","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy1004fju04u54hiadn","texte_question":"

For which values of $$k$$ does $$\\int_{-2}^{k}x^3dx=0$$?

","type_question":"QCM","explication":"

Compute the integral.

Find the antiderivative:

$$\\int x^3dx=\\frac{x^4}{4}.$$

Evaluate from $$-2$$ to $$k$$:

$$\\int_{-2}^{k}x^3dx=\\frac{k^4}{4}-\\frac{(-2)^4}{4}.$$

Since

$$(-2)^4=16,$$

we get

$$\\frac{k^4}{4}-\\frac{16}{4}.$$

Simplify:

$$\\frac{k^4}{4}-4.$$

Set equal to zero:

$$\\frac{k^4}{4}-4=0.$$

Multiply both sides by $$4$$:

$$k^4-16=0.$$

Solve:

$$k^4=16.$$

Take fourth roots:

$$k=\\pm2.$$

Final answer: $$-2$$ and $$2$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-2$$ and $$2$$ ","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-2$$, $$0$$ and $$2$$ ","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy1004kju04j6a2rad4","texte_question":"

If a quantity $$y$$ is directly proportional to its rate of change, which of the following could represent $$y$$ as a function of $$t$$?

","type_question":"QCM","explication":"$23","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$y=5e^{3t}$$ ","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$y=t^4$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$y=\\ln t$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$y=\\frac{1}{t}$$ ","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy1004pju04un6juqz9","texte_question":"

On which interval(s) is the graph of $$y=3x^3-6x^2+2$$ decreasing?

","type_question":"QCM","explication":"

A function is decreasing where its derivative is negative.

Differentiate:

$$\\frac{dy}{dx}=9x^2-12x.$$

Factor:

$$\\frac{dy}{dx}=3x(3x-4).$$

Critical points occur when

$$3x(3x-4)=0.$$

Thus,

$$x=0$$ and $$x=\\frac{4}{3}.$$

Now test intervals.

For $$x<0$$:

Both $$x$$ and $$3x-4$$ are negative.

The product is positive.

The function is increasing.

For $$0<x<\\frac{4}{3}$$:

$$x$$ is positive.

$$3x-4$$ is negative.

The product is negative.

The function is decreasing.

For $$x>\\frac{4}{3}$$:

Both factors are positive.

The product is positive.

The function is increasing.

Therefore, the graph is decreasing on $$\\left(0,\\frac{4}{3}\\right).$$

Final answer: $$(0,\\frac{4}{3})

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$(-\\infty,0)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$(0,\\frac{4}{3})","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$(1,\\infty)$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$(0,1)\\cup(1,\\infty)$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy1004uju042wcd5j7f","texte_question":"

For $$f(x)=\\frac{x^2+4}{x}$$, find the value of $$c$$ in $$[1,4]$$ that satisfies the Mean Value Theorem.

","type_question":"QCM","explication":"

Rewrite the function:

$$f(x)=\\frac{x^2}{x}+\\frac{4}{x}$$

$$f(x)=x+\\frac{4}{x}.$$

Compute the average rate of change on $$[1,4]$$:

$$f(4)=4+\\frac{4}{4}=4+1=5.$$

$$f(1)=1+\\frac{4}{1}=1+4=5.$$

So, the slope of the secant line is:

$$\\frac{f(4)-f(1)}{4-1}=\\frac{5-5}{3}=\\frac{0}{3}=0.$$

Mean Value Theorem requires a $$c$$ in $$ (1,4) $$ such that:

$$f'(c)=0.$$

Differentiate:

$$f'(x)=\\frac{d}{dx}\\left(x+\\frac{4}{x}\\right)=1-\\frac{4}{x^2}.$$

Set equal to $$0$$:

$$1-\\frac{4}{c^2}=0.$$

$$1=\\frac{4}{c^2}.$$

$$c^2=4.$$

$$c=2.$$

Final answer: $$2$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$3$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$4$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy1004zju04mhocs5xw","texte_question":"

A slope field for the differential equation $$\\frac{dy}{dx}=g(x)$$ shows slopes that are zero at $$x=0$$, positive for $$x>0$$, and negative for $$x<0$$. Which function could represent $$g(x)$$?

","type_question":"QCM","explication":"$24","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$g(x)=x$$ ","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$g(x)=x^2$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$g(x)=|x|$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$g(x)=-1$$ ","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy20054ju04jg6efc4z","texte_question":"$25","type_question":"QCM","explication":"$26","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{dy}{dx}=x+y$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{dy}{dx}=xy$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{dy}{dx}=x-y$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{dy}{dx}=x^2+y^2$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy20059ju043mplbqlw","texte_question":"$27","type_question":"QCM","explication":"

The net signed area between two curves on $$[a,b]$$ is the integral of $$\\text{top}-\\text{bottom}$$ from $$a$$ to $$b$$.

Since $$g(x)\\ge f(x)$$ on $$[a,b]$$, the top curve is $$g(x)$$ and the bottom curve is $$f(x)$$.

So, the net signed area is:

$$\\int_a^b(g(x)-f(x))dx.$$

Check the choices:

Choice $$A$$ matches this exactly.

Choice $$B$$ equals $$\\int_b^a(f(x)-g(x))dx=-\\int_a^b(f(x)-g(x))dx=\\int_a^b(g(x)-f(x))dx$$, so it has the same value as $$A$$, but the standard expression over $$[a,b]$$ is written from $$a$$ to $$b$$ with $$g(x)-f(x)$$.

Choice $$C$$ would be negative since $$f(x)-g(x)\\le0$$.

Choice $$D$$ would be negative since reversing limits changes the sign.

Final answer: $$\\int_a^b(g(x)-f(x))dx$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\int_a^b(g(x)-f(x))dx$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\int_b^a(f(x)-g(x))dx$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\int_a^b(f(x)-g(x))dx$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\int_b^a(g(x)-f(x))dx$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy2005eju04b7dqrsys","texte_question":"

On the closed interval $$[2,6]$$, the function $$f$$ is continuous. Suppose $$f(2)=4$$ and $$f(6)=1$$. If $$f'(x)<0$$ and $$f''(x)>0$$ for all $$x\\in(2,6)$$, which of the following could be the value of $$f(4)$$?

","type_question":"QCM","explication":"

Since $$f'(x)<0$$ on $$(2,6)$$, the function is strictly decreasing on $$[2,6]$$.

Therefore,

$$4=f(2)>f(4)>f(6)=1.$$

$$1<f(4)<4.$$

Now use concavity.

Since $$f''(x)>0$$ on $$(2,6)$$, the function is concave up.

For a concave up function, the graph lies below the secant line connecting the endpoints.

The secant line from $$x=2$$ to $$x=6$$ has slope

$$m=\\frac{1-4}{6-2}=\\frac{-3}{4}.$$

Equation of the secant line:

$$L(x)=4+\\frac{-3}{4}(x-2).$$

Evaluate at $$x=4$$:

$$L(4)=4+\\frac{-3}{4}(2)=4-\\frac{6}{4}=4-1.5=2.5.$$

Since the function is concave up,

$$f(4)<2.5.$$

Combine both conditions:

$$1<f(4)<2.5.$$

Final answer: $$2$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$3$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$0$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy2005jju041v4bcf7w","texte_question":"

The water level in a cylindrical tank is decreasing at a rate of $$3$$ inches per minute. The radius of the tank is $$18$$ inches. What is the rate at which the volume of water is changing, in cubic inches per minute, when the volume is $$324\\pi$$ cubic inches?

","type_question":"QCM","explication":"

Volume of a cylinder:

$$V=\\pi r^2 h.$$

Since $$r=18$$,

$$V=\\pi(18)^2h=324\\pi h.$$

Differentiate with respect to time $$t$$:

$$\\frac{dV}{dt}=324\\pi\\frac{dh}{dt}.$$

The water level is decreasing at $$3$$ inches per minute:

$$\\frac{dh}{dt}=-3.$$

Substitute:

$$\\frac{dV}{dt}=324\\pi(-3).$$

$$\\frac{dV}{dt}=-972\\pi.$$

Final answer: $$-972\\pi$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$972\\pi$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-972\\pi$$ ","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-486\\pi$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$486\\pi$$ ","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy2005oju04n39duyon","texte_question":"

If $$f(x)=\\arctan(x^2)$$, find $$f'(1).$$

","type_question":"QCM","explication":"

Differentiate using the chain rule.

If $$f(x)=\\arctan(x^2)$$, then

$$f'(x)=\\frac{1}{1+(x^2)^2}\\cdot 2x.$$

Simplify:

$$f'(x)=\\frac{2x}{1+x^4}.$$

Evaluate at $$x=1$$:

$$f'(1)=\\frac{2(1)}{1+1^4}.$$

$$f'(1)=\\frac{2}{1+1}.$$

$$f'(1)=\\frac{2}{2}.$$

$$f'(1)=1.$$

Final answer: $$1$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{1}{2}$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$1$$ ","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{2}{5}$$ ","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{5}$$ ","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmllmqqy2005tju048adk8jdc","texte_question":"

Beginning at the origin, a particle moves along the positive x-axis with a velocity that remains positive but steadily decreases over time. Which of the following graphs could represent the particle's position $$s(t)$$ as a function of time $$t$$?

","type_question":"QCM","explication":"

The particle begins at the origin, so:

$$s(0)=0.$$

The velocity remains positive, so:

$$s'(t)>0$$

for all $$t.$$ Therefore, the position function is increasing at all times.

The velocity steadily decreases over time, so:

$$s'(t)$$ is decreasing.

That means:

$$s''(t)<0.$$

So, the position function is concave down.

Therefore, the graph must:

Start at the origin

Be increasing

Be concave down

Only choice $$B$$ satisfies all these conditions.

Final answer: A curve starting at the origin that is increasing and concave down

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"A straight line through the origin with constant positive slope","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"A curve starting at the origin that is increasing and concave down ","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"A curve starting above the origin that is increasing and concave up","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"A curve starting at the origin that decreases for some time before increasing ","correcte":false,"explication":"","imageUrl":"$undefined"}]}]},"categories":["Full Practice Exam 1"],"date":"2026-02-13T18:02:44.207Z","modified":"2026-02-20T11:29:30.438Z"}}]]}]}]]