Calculus 2, mathematics answers help

Jan 12th, 2016
Anonymous
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Mathematics
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Question description

1. $\mbox{Find an antiderivative }P(t)\mbox{ of }p(t)=t^5-\frac{t^3}{3}-t.$
$\frac{t^7}{6}-\frac{t^5}{12}-\frac{t^2}{2}+C$
$\frac{t^6}{6}-\frac{t^4}{12}-\frac{t^2}{2}+C$
$\frac{t^6}{3}-\frac{t^4}{24}-\frac{t^2}{2}+C$
$t^6-\frac{t^4}{3}-t^2+C$
2. $\mbox{If }f(x)=6x^8-3x^5-7x^3+4x,\mbox{ find }f\prime(x).$
$48x^7-3x^4-7x^2+4$
$48x^7-15x^4-21x^2+4$
$48x^7-15x^4-21x^2+4x$
$6x^7-15x^4-21x^2+4$
3. $\mbox{Evaluate the indefinite integral }$$\int {\hat{A}}{\mbox{ sin}}^{8}(7x)\mbox{ cos }(7x)dx.$
$\frac{1}{9}{\mbox{sin}}^{9}(7x)+C$
$\frac{1}{63}{\mbox{sin}}^{9}(7x)+C$
$\frac{1}{63}{\mbox{cos}}^{9}(7x)+C$
${\mbox{sin}}^{9}(7x)+C$
4. $\mbox{For the equation }{(3x-y)}^{4}+4y^4=2405,$$\mbox{ evaluate }\frac{dy}{dx}\mbox{ at the point }(-2,1).$
$\frac{1029}{170}$
$\frac{1029}{340}$
$\frac{49}{340}$
$\frac{343}{340}$
5. $\mbox{If }f(x)=2x\sqrt{x}+\frac{5}{x^2{\sqrt{x}}},\mbox{ find }f\prime(4).$
$\frac{1511}{256}$
$\frac{1387}{256}$
$\frac{1571}{256}$
$\frac{1253}{256}$
6. $\mbox{Evaluate }$$\underset{x\rightarrow  \infty }{\mbox{lim}}\left (\sqrt{x^2+5x+1}-x \right ).$$\mbox{ Hint: rationalizing the numerator.}$
$ \infty $
$0$
$\frac{5}{2}$
$5$
$\frac{7}{2}$
7. $\mbox{Evaluate the indefinite integral }$$\int \frac{25x}{{(1-5x^{2})}^2}dx.$
$\frac{25}{1-5x^2}+C$
$\frac{25}{2{(1-5x^{2})}^2}+C$
$\frac{5}{2(1-5x^{2})}+C$
$\frac{1}{1-5x^2}+C$
8. $\mbox{Evaluate the limit }$$\underset{h\rightarrow 0}{\mbox{lim}}\frac{8{(a+h)}^2-8a^2}{h}.$
$16a$
$\mbox{Does not exist}$
$a^2$
$8a$
$16a^2$
9. $\mbox{Evaluate the following limits. }$
$\mbox{(a) }\underset{x\rightarrow \infty }{\mbox{lim}}\frac{6x+5}{8x^2-7x+6}$
$\mbox{(b) }\underset{x\rightarrow {-\infty}}{\mbox{lim}}\frac{6x+5}{8x^2-7x+6}$
$\mbox{(a) }0 \mbox{ and (b) }0 $
$\mbox{(a) }\frac{3}{4}\mbox{ and (b) }-\frac{3}{4}$
$\mbox{(a) } \infty \mbox{ and (b) }- \infty $
$\mbox{(a) }0\mbox{ and (b) } -\infty $
10. $\mbox{Use implicit differentiation to find the slope of the tangent line to the curve }$$\frac{y}{x-7y}=x^6-6\mbox{ at the point } \left ( 1,\frac{5}{34} \right ) .$
$\frac{44}{289}$
$\frac{63}{34}$
$\frac{54}{289}$
$\frac{23}{289}$
11. $\mbox{Evaluate the limit using l'Hôpital's rule if necessary. }$$\underset{x\rightarrow \infty}{\mbox{lim}}{\left (1+\frac{3}{x} \right )}^{\frac{x}{11}}.$
$e^{11}$
$0$
$e^3$
$e^{\frac{3}{11}}$
$\mbox{Does not exist}$
12. $\mbox{Let }f(x)={(\mbox{ln }x)}^{8}.\mbox{ Find }f\prime(e^2).$
$\frac{128}{e^2}$
$\frac{512}{e^2}$
$\frac{1024}{e^2}$
$\frac{256}{e^2}$
13. $\mbox{Evaluate exactly, using the Fundamental Theorem of Calculus: }$$\int_{0}^{s} \left (\frac{x^6}{11}+2x \right ) dx.$
$\frac{s^7}{77}+2s^2$
$\frac{s^7}{11}+s^2$
$\frac{s^7}{77}+s^2$
$\frac{s^7}{77}+s^3$
14. $\mbox{Let }f(x)=x^{3}{\mbox{tan }}^{-1}(2x).\mbox{ Find }f\prime(x).$
$\frac{x^3}{1+2x^2}+3x^{2}{\mbox{tan }}^{-1}(2x)$
$\frac{x^3}{1+2x^2}+{\mbox{tan }}^{-1}(2x)$
$\frac{1}{1+4x^2}+3x^{2}{\mbox{tan }}^{-1}(2x)$
$\frac{x^3}{1+4x^2}+3x^{2}{\mbox{tan }}^{-1}(2x)$
15. $\mbox{Find the derivative of the function }f(x)=\mbox{arctan }(x^9).$$\mbox{ Hint: It may be to your advantage to simplify before differentiating.}$
$\frac{9x^8}{1+x^{9}}$
$\frac{x^8}{1+x^{18}}$
$\frac{9x^8}{1+x^{18}}$
$\frac{1}{1+x^{18}}$
16. $\mbox{Evaluate the limit }$$\underset{x\rightarrow 0}{\mbox{lim}}\frac{\mbox{tan }9x}{\mbox{sin }3x}.$
$3$
$2$
$0$
$\mbox{Does not exist}$
17. $\mbox{Find the derivative of }f(x)=x^{2}\mbox{cos }x.$
$x^{2}\mbox{sin }x+2x\mbox{ cos }x$
$-x^{2}\mbox{sin }x+x\mbox{ cos }x$
$-x^{2}\mbox{sin }x+2x\mbox{ cos }x$
$x^{2}\mbox{sin }x+x\mbox{ cos }x$
18. $\mbox{Evaluate the limit }$$\underset{x\rightarrow {-1}}{\mbox{lim}}\frac{x^2+10x+9}{x+1}.$
$0$
$\mbox{Does not exist}$
$8$
$4$
19. $\mbox{Evaluate the limit using l'Hôpital's rule. }$$\underset{x\rightarrow {1}}{\mbox{lim}}\frac{x^{14}-1}{x^3-1}.$
$0$
$-\frac{7}{3}$
$\frac{14}{3}$
$\frac{7}{3}$
$\mbox{Does not exist}$
20. $\mbox{Consider the function }f(x)=2x^2-2x^3+4x.$$\mbox{ Find the average slope }m\mbox{ of this function on the interval }(-2,4).$$\mbox{ By the Mean Value Theorem, we know there exists }$$\mbox{a }c\mbox{ in the open interval }(-2,4)$$\mbox{ such that }f\prime(c)\mbox{ is equal to this mean slope. }$$\mbox{Find the two values of }c\mbox{ in the interval which works.}$
$m=-16, c=2.19\mbox{ or }-1.52$
$m=-\frac{40}{3}, c=2.19\mbox{ or }-1.52$
$m=-\frac{40}{3}, c=2.54\mbox{ or }-1.72$
$m=-16, c=2.54\mbox{ or }-1.72$


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