Mini Exam 6
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In the $$xy$$-plane, the function $$f$$ is graphed and contains the point $$(4,3)$$. Which of the following points is on the graph of $$y=3f(x)-2$$?⌄
The point $$(4,3)$$ means $$f(4)=3$$Use the transformation.$$y=3f(x)-2$$Substitute$$x=4$$. $$y=3f(4)-2$$$$y=3(3)-2$$ $$y=9-2$$$$y=7$$ So the point is $$(4,7)$$Final answer: $$(4,7)$$
Consider the piecewise function$$P(t)=200-0.5t^2$$, $$0\le t$$<$$10$$$$P(t)=150-5t$$, $$10\le t\le12$$.Which of the following statements is correct about the average rates of change of $$P$$?⌄
Compute average rate on $$0\le t\le5$$.$$P(0)=200$$$$P(5)=200-0.5(25)=187.5$$$$\frac{187.5-200}{5-0}=-2.5$$Compute average rate on $$5\le t\le10$$.$$P(10)=200-0.5(100)=150$$$$\frac{150-187.5}{10-5}=-7.5$$Compare.$$-2.5$$>$$-7.5$$So, the rate on $$0\le t\le5$$ is greater.Final answer: $$\text{A}$$
A table of values for a function $$A(x)$$ is shown. A quadratic model is used to fit the data. Which of the following is the best estimate of the maximum value of $$A(x)$$?⌄
From the table.$$A(10)=220$$$$A(12)=260$$$$A(15)=295$$$$A(18)=280$$The values increase then decrease.So, the maximum is near $$x=15$$.The maximum value is approximately $$295$$Final answer: $$295$$
The function $$M$$ is defined by $$M(t)=ae^{-0.02t}+b$$, where $$a$$ and $$b$$ are positive constants. If $$M(0)=20$$ and $$\lim_{t\to\infty}M(t)=4$$, for what value of $$t$$ is $$M(t)=10$$?⌄
Find $$b$$. $$\lim_{t\to\infty}M(t)=b$$ So $$b=4$$ Use $$M(0)=20$$. $$a+4=20$$ $$a=16$$ Write the function. $$M(t)=16e^{-0.02t}+4$$ Set equal to $$10$$. $$16e^{-0.02t}+4=10$$ $$16e^{-0.02t}=6$$ $$e^{-0.02t}=\frac{3}{8}$$ Take natural log. $$-0.02t=\ln\left(\frac{3}{8}\right)$$ $$t=\frac{-\ln\left(\frac{3}{8}\right)}{0.02}$$ $$t\approx48.96$$ Final answer: $$48.96$$
Which of the following intervals represents all $$x$$ such that $$p(x)\ge0$$ for $$p(x)=x(x-3)(x+1)$$?⌄
Find the zeros. $$x=-1$$ $$x=0$$ $$x=3$$ Test the interval $$(-\infty,-1)$$. Using $$x=-2$$, $$(-2)(-5)(-1)$$<$$0$$ Test the interval $$(-1,0)$$. Using $$x=-\frac{1}{2}$$, $$\left(-\frac{1}{2}\right)\left(-\frac{7}{2}\right)\left(\frac{1}{2}\right)$$>$$0$$ Test the interval $$(0,3)$$. Using $$x=1$$, $$(1)(-2)(2)$$<$$0$$ Test the interval $$(3,\infty)$$. Using $$x=4$$, $$(4)(1)(5)$$>$$0$$ Include the zeros because $$p(x)\ge0$$. The solution set is $$[-1,0]\cup[3,\infty)$$ Final answer: $$[-1,0]\cup[3,\infty)$$
The function $$f$$ is given by $$f(x)=4\cdot16^x$$. Which of the following is an equivalent form of $$f(x)$$?⌄
Rewrite $$16$$ as a power of $$2$$. $$16=2^4$$ Substitute. $$f(x)=4\cdot(2^4)^x$$ Use the power rule. $$f(x)=4\cdot2^{4x}$$ Final answer: $$4\cdot2^{4x}$$
The function $$f$$ is given by $$f(x)=-3x^5+2x^2-1$$. Which of the following describes the end behavior of $$f$$?⌄
The leading term is $$-3x^5$$ The degree is odd. The leading coefficient is negative. So, as $$x\to\infty$$, $$f(x)\to-\infty$$ So, as $$x\to-\infty$$, $$f(x)\to\infty$$ Final answer: $$\lim_{x\to-\infty}f(x)=\infty\text{ and }\lim_{x\to\infty}f(x)=-\infty$$
Let $$a$$ and $$b$$ be positive constants. Which of the following is equivalent to $$3\ln a-2\ln b$$?⌄
Use the power rule. $$3\ln a=\ln(a^3)$$ $$2\ln b=\ln(b^2)$$ Substitute. $$3\ln a-2\ln b=\ln(a^3)-\ln(b^2)$$ Use the quotient rule. $$\ln(a^3)-\ln(b^2)=\ln\left(\frac{a^3}{b^2}\right)$$ Final answer: $$\ln\left(\frac{a^3}{b^2}\right)$$