Ap calculus Definition and 117 Threads

I originally thought you’d have to use the chain rule to get h’, as in: f’(g(x))*g’(x). Plugging in 1 for x, I got an answer of 10. An online solution, however, said that you only had to get f(g(1)), which was f(-1), then look up f’(-1) in the table. Both approaches seem logical to me, but they...

screen shot to avoid typos OK the key said it was D I surfed for about half hour trying to find a solution to this but $f'(0)$ doesn't equal any of these numbers $e^0=\pm 1$ from the $e^{(x^2-1)^2}$ kinda ?

ok I thot this was just observation to get b. but maybe not I saw some rather hefty substations to get different answers

$\tiny{4.2.5}$ $\displaystyle\int^1_0{xe^x\ dx}$ is equal to $A.\ \ {1}\quad B. \ \ {-1}\quad C. \ \ {2-e}\quad D.\ \ {\dfrac{e^2}{2}}\quad E.\ \ {e-1}$ ok I think this is ok possible typos but curious if this could be solve not using IBP since the only variable is x

$\displaystyle\lim_{x \to 0}\dfrac{1-\cos^2(2x)}{(2x)^2}=$ by quick observation it is seen that this will go to $\dfrac{0}{0)}$ so L'H rule becomes the tool to use but first steps were illusive the calculator returned 1 for the Limit

1. $f(x)=(2x+1)^3$ and let g be the inverse function of f. Given that$f(0)=1$ what is the value of $g'(1)$? A $-\dfrac{2}{27}$ B $\dfrac{1}{54}$ C $\dfrac{1}{27}$ D $\dfrac{1}{6}$ E 6 2. given that $\left[f(x)=x-2,\quad g(x)=\dfrac{x}{x^2+1}\right]$ find $f(g(-2))$...

by observation I choose (c) since the limit values may not be =

I thot I posted this before but couldn't find it ... if so apologize Let f be a function that is continuous on the closed interval [2,4] with $f(2)=10$ and $f(4)=20$ Which of the following is guaranteed by the $\textbf{Intermediate Value Theorem?}$ $a.\quad f(x)=13 \textit{ has a least one...

Evaluate $\displaystyle\int{\dfrac{{(1-\ln{t})}^2}{t} dt=}$ $a\quad {-\dfrac{1}{3}{(1-\ln{t})}^3+C} \\$ $b\quad {\ln{t}-2\ln{t^2} +\ln{t^3} +C} \\$ $c\quad {-2(1-\ln{t})+C} \\$ $d\quad {\ln{t}-\ln{t^2}+\dfrac{(\ln{t^3})}{3}+C} \\$ $e\quad {-\dfrac{(1-\ln{t^3})}{3}+C}$ ok we can either expand...

$\textsf{What is the area of the region in the first quadrant bounded by the graph of}$ $$y=e^{x/2} \textit{ and the line } x=2$$ a. 2e-2 b. 2e c. $\dfrac{e}{2}-1$ d. $\dfrac{e-1}{2}$ e. e-1Integrate $\displaystyle \int e^{x/2}=2e^{x/2}$ take the limits...

Ok I thot I posted this before but after a major hunt no find Was ? With these options since if f(x) Is a curve going below the x-axis Is possible and () vs [] If this is a duplicate post What is the link.. I normally bookmark these Mahalo ahead

If $f^{-1}(x)$ is the inverse of $f(x)=e^{2x}$, then $f^{-1}(x)=$$a. \ln\dfrac{2}{x}$ $b. \ln \dfrac{x}{2}$ $c. \dfrac{1}{2}\ln x$ $d. \sqrt{\ln x}$ $e. \ln(2-x)$ ok, it looks slam dunk but also kinda ? my initial step was $y=e^x$ inverse $\displaystyle x=e^y$ isolate $\ln{x} = y$ the...

ok well it isn't just adding the areas of 2 functions but is $xf(x)$ as an integrand Yahoo had an answer to this but its never in Latex so I couldn't understand how they got $\dfrac{7}{2}$

https://www.physicsforums.com/attachments/9542 ok all I did was count squares so about 21

$\displaystyle g'=2xe^{kx}+e^{kx}kx^2$ we are given $ x=\dfrac{2}{3}$ then $\displaystyle g'=\dfrac{4}{3}e^\left(\dfrac{2k}{3}\right)+e^\left({\dfrac{2k}{3}}\right)\dfrac{4k}{9}$ ok something is ? aren't dx supposed to set this to 0 to find the critical point did a desmos look like k=-3 but ...

https://www.physicsforums.com/attachments/9527 ok from online computer I got this $\displaystyle\int_0^x e^{-t^2}=\frac{\sqrt{\pi }}{2}\text{erf}\left(t\right)+C$ not sure what erf(t) means

ok I chose e for the zeros

OK, this can only be done by observation so since we have v(t) I chose e but the eq should have a minus sign. here the WIP version of the AP Calculus Exam PDF as created in Overleaf https://documentcloud.adobe.com/link/track?uri=urn%3Aaaid%3Ascds%3AUS%3A053a75d8-ca5b-4447-bd65-4e580f0de793...

ok I got stuck real soon... .a find where the functions meet $$\ln x = 5-x$$ e both sides $$x=e^{5-x}$$ok how do you isolate x? W|A returned $x \approx 3.69344135896065...$ but not sure how they got itb.? c.?

yes I know this is a very common problem but likewise many ways to solve it ok I really have a hard time with these took me 2 hours to do this looked at some examples but some had 3 variables and 10 steps confusing to get the ratios set up... ok my take on it is here see if you can solve...

ok I chose c being false, since a limit does not exist if f(x) is different coming from $\pm$

image due to graph, I tried to duplicate this sin wave on desmos but was not able to. so with sin and cos it just switches to back and forth for the derivatives so thot a this could be done just by observation but doesn't the graph move by the transformations well anyway?

ok this one baffled me a little but isn't $g = e^{-t^2}$and the graph of that has an inflection point at x=1

A particle moves along the x-axis. The velocity of the particle at time t is $6t - t^2$. What is the total distance traveled by the particle from time $t = 0$ to $t = 3$ ok we are given $v(t)$ so we do not have to derive it from a(t) since the initial $t=0$ we just plug in the $t=3$ into $v(t)$...

I tried to do this just by observation, but kinda hard with a piece wise function so would presume

ok these always baffle me because f(t) is not known. however if $f'(t)>0$ then that means the slope is aways positive which could be just a line. but could not picture this to work in the tables. Im sure the answer can be found quickly online but I don't learn by copy and paste. d was...

image due to macros in Overleaf ok I think (a) could just be done by observation by just adding up obvious areas but (b) and (c) are a litte ? sorry had to post this before the lab closes

image due to macros in overleaf well apparently all we can do is solve this by observation which would be the slope as x moves in the positive direction e appears to be the only interval where the slope is always increasing

If $f(x)=7x-3+\ln(x),$ then $f'(1)=$ $a.4\quad b. 5\quad c. 6\quad d. 7\quad e. 8$ see if you can solve this before see the proposed solution

ok image to avoid typo... try to solve before looking at suggested solutions ok I think you could do this by observation if you are careful with signs

The graph of $y=e^{\tan x} - 2$ crosses the x-axis at one point in the interval [0,1]. What is the slope of the graph at this point. A. 0.606 B 2 C 2.242 D 2.961 E 3.747ok i tried to do a simple graph of y= with tikx but after an hour trying failed doing this in demos it seens the answer is...

309 average temperature $$\begin{array}{|c|c|c|c|c|c|c|} \hline t\,(minutes)&0&4&9&15&20\\ \hline W(t)\,(degrees Farrenheit)&55.0&57.1&61.8&67.9&71.0\\ \hline \end{array}$$ The temperature of water in a tub at time t is modeled by a strictly increasing, twice-differentiable function W. where...

ok basically t is 3 hours appart except between 7 and 12 of which I didn't know if we should intemperate. other wise it is just adding up the 4 $(t)\cdot(R(t))$s.

ok just posted an image due to macros in the overleaf doc this of course looks like a sin or cos wave and flips back and forth by taking derivatives looks like a period of 12 and an amplitude of 3 so... but to start I was not able to duplicate this on desmos altho I think by observation alone...

ok again I used an image since there are macros and image I know this is a very common problem in calculus but think most still stumble over it inserted the graph of v(t) and v'(t) and think for v'(t) when the graph is below the x-axis that participle is moving to the left the integral has a...

ok I posted a image to avoid any typos but was wondering why the question has dx and options are in dt I picked C from observation but again that was assuming f was a horizontal line of which it could be something else that way the limits stay the same but the area is cut in halfopinions...

I just posted a image due to overleaf newcommands and graph ok (a) if we use f(20) then the $B=0$ so their no weight gain. (b), (c), was a little baffled and not sure how this graph was derived...

The function f is defined by $$f(x)=\sqrt{25-x^2},\quad -5\le x \le 5$$ (a) Find $f'(x)$ apply chain rule $$ \dfrac{d}{dx}(25-x^2)^{1/2} =\dfrac{1}{2}(25-x^2)^{-1/2}2x =-\frac{x}{\sqrt{25-x^2}}$$ (b) Write an equation for the tangent line to the graph of f at $x=-3$...

ok probably if one did a lot of these this could be solved by observation others separate the variables and take to Integral to get the equation

I'm just going to post this image now since my tablet won't render the latex. This is a free response question.. But my experience is that the methods of solving are more focused here at mhb saving many error prone steps.. Mahalo ahead...

calculator returned this but know sure why $\displaystyle2 \int _1^2e^udu$ note there might be a duplicat of this post ?

Ok not sure if I fully understand the steps but presume the first step would be divide both sides deriving$$\dfrac{dy}{dx}=\dfrac{2x-y}{x+2y}$$offhand don't know the correct answer $\tiny{from College Board}$

Let $g(x)$ be the function given by $g(x) = x^2e^{kx}$ , where k is a constant. For what value of k does g have a critical point at $x=\dfrac{2}{3}$? $$(A)\quad {-3} \quad (B)\quad -\dfrac{3}{2} \quad (C)\quad -\dfrac{3}{2} \quad (D)\quad {0} \quad (E)\text{ There is no such k}$$ ok I...

238 Solve $$\displaystyle\lim_{h\to 0} \dfrac{\ln{(4+h)}-\ln{h}}{h}$$ $$(A)\,0\quad (B)\, \dfrac{1}{4}\quad (C)\, 1\quad (D)\, e\quad (E)\, DNE$$ The Limit diverges so the Limit Does Not Exist (E)ok the only way I saw that it diverges is by plotting not sure what the rule is that observation...

AP Calculas Exam Problem$\textsf{Using $\displaystyle u=\sqrt{x}, \quad \int_1^4\dfrac{e^{\sqrt{x}}}{\sqrt{x}}\, dx$ is equal to which of the following}$ $$ (A)2\int_1^{16} e^u \, du\quad (B)2\int_1^{4} e^u \, du\quad (C) 2\int_1^{2} e^u \, du\quad (D) \dfrac{1}{2}\int_1^{2} e^u \, du\quad...

$\tiny{237}$ $\textsf{The total area of the region bounded by the graph of $f(x)=x\sqrt{1-x^2}$ and the x-axis is}$ $$(A) \dfrac{1}{3}\quad (B) \dfrac{1}{3}\sqrt{2}\quad (C) \dfrac{1}{2}\quad (D) \dfrac{2}{3}\quad (E) 1 $$ find the limits of integration if $$f(x)=0 \textit{...

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Let $f(x)=(2x+1)^3$ and let g be the inverse of $f$. Given that $f(0)=1$, what is the value of $g'(1)?$$(A)\, \dfrac{2}{27} \quad (B)\, \dfrac{1}{54} \quad (C)\, \dfrac{1}{27} \quad (D)\, \dfrac{1}{6} \quad (E)\, 6$ok not sure what the best steps on this would be but assume we first find...

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