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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #12 : Calculus I — Derivatives

Define $g(x) = x^{3} -9x^{2}+24x+17$ .

Give the interval(s) on which is decreasing.

Possible Answers:

(2,4)

$(3,\infty)$

$(-\infty ,3)$

$(-\infty ,2) \cup (4,\infty)$

$(-\infty ,\infty)$

Correct answer:

(2,4)

Explanation:

is decreasing on those intervals at which g'(x)<0 .

$g(x) = x^{3} -9x^{2}+24x+17$

$g'(x) = 3\cdot x^{3-1} -2\cdot 9x^{2-1}+24+0$

$g'(x) = 3x^{2} -18x+24$

We need to find the values of for which $g'(x) = 3x^{2} -18x+24 <0$ . To that end, we first solve the equation:

$3x^{2} -18x+24 =0$

$3 \left (x^{2} -6x+8 \right ) =0$

$3 \left (x-2 \right )\left (x-4 \right ) =0$

$x=2\textrm{ or } x = 4$

These are the boundary points, so the intervals we need to check are:

$(-\infty ,2)$ , (2,4) , and $(4,\infty)$

We check each interval by substituting an arbitrary value from each for .

$(-\infty ,2)$

Choose x = 0

$3x^{2} -18x+24 =3\cdot 0^{2} -18\cdot 0+24 =24 > 0$

increases on this interval.

(2,4)

Choose x = 3

$3x^{2} -18x+24 =3\cdot 3^{2} -18\cdot 3+24 =27 -54 +24 = -3 < 0$

decreases on this interval.

$(4,\infty)$

Choose x = 5

$3x^{2} -18x+24 =3\cdot 5^{2} -18\cdot 5+24 = 75 -90 +24 = 9 > 0$

increases on this interval.

The answer is that decreases on (2,4) .

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Example Question #13 : Calculus I — Derivatives

Define $f \left ( x\right ) = x^{3} - \frac{21}{2} x^{2} +30x -6$ .

Give the interval(s) on which is increasing.

Possible Answers:

$(-\infty ,\infty)$

$(-\infty ,2) \cup (5,\infty)$

(2,5)

$\left ( -\infty , 3 \frac{1}{2}\right )$

$\left ( 3 \frac{1}{2}, \infty \right )$

Correct answer:

$(-\infty ,2) \cup (5,\infty)$

Explanation:

is increasing on those intervals at which f'(x)>0 .

$f \left ( x\right ) = x^{3} - \frac{21}{2} x^{2} +30x -6$

$f' \left ( x\right ) = 3x^{3-1} - 2 \cdot \frac{21}{2} x^{2-1} +30 -0$

$f' \left ( x\right ) = 3x^{2} - 21 x +30$

We need to find the values of for which $f' \left ( x\right ) = 3x^{2} - 21 x +30 > 0$ . To that end, we first solve the equation:

$3x^{2} - 21 x +30 = 0$

$3(x^{2} - 7 x +10 )= 0$

3(x-2)(x-5)= 0

$x = 2 \textrm{ or }x = 5$

These are the boundary points, so the intervals we need to check are:

$(-\infty ,2)$ , (2,5) , and $(5,\infty)$

We check each interval by substituting an arbitrary value from each for .

$(-\infty ,2)$

Choose x = 0

$3x^{2} - 21 x +30 = 3 \cdot 0^{2} - 21 \cdot 0 +30 = 30 > 0$

increases on this interval.

(2,5)

Choose x = 3

$3x^{2} - 21 x +30 = 3 \cdot 3^{2} - 21 \cdot 3 +30 = 27 -63 +30 = -6 < 0$

decreases on this interval.

$(5,\infty)$

Choose x = 6

$3x^{2} - 21 x +30 = 3 \cdot 6^{2} - 21 \cdot 6 +30 = 108 -126 + 30=12 > 0$

increases on this interval.

The answer is that increases on $(-\infty ,2) \cup (5,\infty)$

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Example Question #3 : Derivatives

At what point does -8x^2+15 shift from increasing to decreasing?

Possible Answers:

It does not shift from increasing to decreasing

x=-8

x=-16

$x=-\frac{1}{16}$

x=0

Correct answer:

x=0

Explanation:

To find out where the graph shifts from increasing to decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as 15x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{2-1})+(0*15x^{0-1})$

Notice that $(0*15x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

If we were to graph y=-16x , would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

y=-16x

0=-16x

$\frac{0}{-16}=x$

0=x

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Example Question #1 : Derivatives

At what value of does 5x^2+12x shift from decreasing to increasing?

Possible Answers:

$x=-\frac{6}{5}$

x=-12

$x=\frac{5}{6}$

x=-10

It does not shift from decreasing to increasing

Correct answer:

$x=-\frac{6}{5}$

Explanation:

To find out when the function shifts from decreasing to increasing, we look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{1})+(1*12x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

From here, we want to know if there is a point at which graph changes from negative to positive. Plug in zero for y and solve for x.

y=10x+12

0=10x+12

-12=10x

$\frac{-12}{10}=x$

$\frac{-6}{5}=x$

This is the point where the graph shifts from decreasing to increasing.

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Example Question #1 : Understanding L'hospital's Rule

Evaluate the following limit:

$\lim_{x\rightarrow 0} (\frac{1-cos2x}{sinx})$

Possible Answers:

$\infty$

Correct answer:

Explanation:

When approaches 0 both sinx and (1-cos2x) will approach . Therefore, L’Hopital’s Rule can be applied here. Take the derivatives of the numerator and denominator and try the limit again:

$\lim_{x\rightarrow 0} (\frac{1-cos2x}{sinx})= \lim_{x\rightarrow 0} (\frac{2sin2x}{cosx})=\frac{0}{1}=0$

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Example Question #2 : Derivative Interpreted As An Instantaneous Rate Of Change

Find the instantaneous rate of change for the function,

y = 2x^3 - x-4

at the point (-1,-7) .

Possible Answers:

293

712

Correct answer:

Explanation:

Find the instantaneous rate of change for the function,

y = 2x^3 - x-4

at the point (-1,-7) .

1) First compute the derivative of the function, since this will give us the instantaneous rate of change of the function as a function of .

$\frac{dy}{dx}=\frac{d}{dx}(2x^3-x-4)$

$\frac{dy}{dx}=\frac{d}{dx}(2x^3)-\frac{d}{dx}(x)-\frac{d}{dx}(4)$

$\frac{dy}{dx}= 6x^2-1$

2) Now evaluate the derivative at the value x = -1 ,

$\frac{dy}{dx}_{|x=-1}= 6(-1)^2-1 = 5$

Therefore, is the instantaneous rate of change of the function

y = 2x^3 - x-4 at the point (-1,-7) .

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Example Question #3 : Derivative Interpreted As An Instantaneous Rate Of Change

A particle is traveling in a straight line along the x-axis with position function $x(t) = t+\sin(t)$ . What is the instantaneous rate of change in the particle's position at time $\pi$ seconds?

Possible Answers:

$\pi+1$

$\pi$

$\pi-1$

Correct answer:

Explanation:

To find the instantaneous rate of change of the particle at time $t=\pi$ , we have to find the derivative of x(t) and plug $\pi$ into it.

$x'(t) :=v(t) = 1+\cos(t)$ .

And

$v(\pi) = 1+\cos(\pi) = 1+-1 = 0$ .

Hence the instantaneous rate of change in position (or just 'velocity') of the particle at $t=\pi$ is . (At that very instant, the particle is not moving.)

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Example Question #4 : Derivative Interpreted As An Instantaneous Rate Of Change

Find the function values and as well as the instantaneous rate of change for the function $g(x)=x\sin(x)$ corresponding to the following values of

$1)\: \: \: \: \: \: \: x =\frac{\pi}{2}$

$2)\: \: \: \: \: \: \: x =\frac{\pi}{4}$

$3)\: \: \: \: \: \: \: x =\pi$

Possible Answers:

B2wrng4

Fixedcorrectanswer

B2wrng3

B2wrng4

Correct answer:

Fixedcorrectanswer

Explanation:

Find the instantaneous rate of change for the function $g(x)=x\sin(x)$ corresponding to the following values of

$1)\: \: \: \: \: \: \: x =\frac{\pi}{2}$

$2)\: \: \: \: \: \: \: x =\frac{\pi}{4}$

$3)\: \: \: \: \: \: \: x =\pi$

Evaluate the function at each value of

$g\left(\frac{\pi}{2} \right )=\frac{\pi}{2}sin\left(\frac{\pi}{2} \right )=\frac{\pi}{2}$

$g\left(\frac{\pi}{4} \right )=\frac{\pi}{4}\sin\left(\frac{\pi}{4} \right )=\frac{\pi}{4}\frac{\sqrt{2}}{2}=\frac{\pi\sqrt{2}}{8}$

$g(\pi)=\pi \sin(\pi)=0$

The instantaneous rate of change at any point (x, g(x)) will be given by the derivative at that point. First compute the derivative of the function:

$g(x)=x\sin(x)$

Apply the product rule:

$g'(x)=x\frac{d}{dx}\sin(x)+\sin(x)\frac{d}{dx}(x)$

$= x\cos(x)+\sin(x)$

Therefore,

$g'(x)=x\cos(x)+\sin(x)$

Now evaluate the derivative for each given value of :

$1)\: \: \: \: \: \: \: x =\frac{\pi}{2}\: \: \: \:\rightarrow\: \: \: \:g'\left(\frac{\pi}{2} \right )\: \: =\: \: \frac{\pi}{2}\cos\left(\frac{\pi}{2} \right )+\sin\left(\frac{\pi}{2} \right )\: \:=\:1$

$2)\: \: \: \: \: \: \: x =\frac{\pi}{4}\: \: \: \:\rightarrow\: \: \: \:g'\left(\frac{\pi}{4} \right )\: \: =\: \: \frac{\pi}{4}\cos\left(\frac{\pi}{4} \right )+\sin\left(\frac{\pi}{4} \right )\: \:=\:\frac{\sqrt{2}}{2}\left(1+\frac{\pi}{4} \right )=\frac{\sqrt{2}}{8}(4+\pi)$

$3)\: \: \: \: \: \: \: x =\pi\: \: \: \:\rightarrow\: \: \: \:g'(\pi)\: \: =\: \: \pi\cos(\pi) +\sin(\pi) \: \:=\:\pi$

Therefore, the instantaneous rate of change of the function g(x) at the corresponding values of are:

Fixedcorrectanswer

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Example Question #5 : Derivative Interpreted As An Instantaneous Rate Of Change

Given that v(t) is the velocity of a particle, find the particle's acceleration when t=3.

v(t)=3t^2+6e^x-15t

Possible Answers:

123.5

12.65

Not enough information provided

Correct answer:

123.5

Explanation:

Given that v(t) is the velocity of a particle, find the particle's acceleration when t=3.

v(t)=3t^2+6e^x-15t

We are given velocity and asked to find acceleration. Our first step should be to find the derivative.

v(t)=3t^2+6e^x-15t

We can use our standard power rule for our 1st and 3rd terms, but we need to remember something else for our second term. Namely, that the derivative of e^x is simply e^x

With that in mind, let;s find v'(t)

v(t)=3t^2+6e^x-15t

v'(t)=6t+6e^x-15

Now, for the final push, we need to find the acceleration when t=3. We do this by plugging in 3 for t and simplifying.

v'(t)=6(3)+6e^3-15=18+120.5-15=123.5

So, our answer is 123.5

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Example Question #6 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\\&p(t)=6t - sin(\frac{(5\cdot \pi t)}{2}) + 5\\&\text{What is its velocity at time }t=6\text{?}\end{align*}$

Possible Answers:

6.83

13.85

Correct answer:

13.85

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function }6t - sin(\frac{(5\cdot \pi t)}{2}) + 5\\&\text{We find the velocity function: }6 -\frac{ (5\cdot \pi cos(\frac{(5\cdot \pi t)}{2}))}{2}\\&\text{Plugging in values we then get:}\\&v(6)=6 -\frac{ (5\cdot \pi cos(\frac{(5\cdot \pi (6))}{2}))}{2}\\&v(6)=13.85\end{align*}$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #12 : Calculus I — Derivatives

Example Question #13 : Calculus I — Derivatives

Example Question #3 : Derivatives

Example Question #1 : Derivatives

Example Question #1 : Understanding L'hospital's Rule

Example Question #2 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #3 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #4 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #5 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #6 : Derivative Interpreted As An Instantaneous Rate Of Change

All AP Calculus AB Resources

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