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AP Calculus BC : Derivatives

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Example Questions

Example Question #1 : Finding Maximums

What are the -coordinate of the local maximum on the graph of the function

p(x)=x^3-27x+12.5 ?

Possible Answers:

x=55

x=3

x=-3

x=9

Correct answer:

x=-3

Explanation:

To find maxima and minima, find the coordinates of the points where the derivative is undefined or equal to zero. The derivative of p(x) is

p'(x)=3x^2-27

Next set the derivative equal to zero and solve for x:

0=3x^2-27

27=3x^2

9=x^2

$\pm3=x$

Finally we need to test the critical points in the original equation to determine which is a maximum.

p(3)=3^3-27(3)+1=-53

p(-3)=(-3)^3-27(-3)+1=55

Since the value of the function is greatest at x = -3, that is the x-coordinate of the maximum.

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Example Question #71 : Derivatives

Find the local maximum of the function $y=\frac{1}{x^2+1}$ .

Possible Answers:

(1,0)

There are none.

(0,1)

(0,0)

Correct answer:

(0,1)

Explanation:

To find the local maximum, first find the first derivative of the function.

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

Then find all values of x for which the derivative equals 0 or is undefined. The derivative equals 0 when x=0 and is never undefined because the denominator is always greater than 0. Then, by picking points less than and greater than 0, we see that the function is increasing less than 0 and increasing greater than 0.

Therefore, it is a local maximum.

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Example Question #1 : Finding Maximums

Find the local maxima of the following function:

$f(x)=\frac{x^3}{3}-8x^2+64x$

Possible Answers:

x=-8

$x=\infty$

There are no local maxima

x=8

Correct answer:

There are no local maxima

Explanation:

To find the local maximum of the function, we must find the point at which the first derivative changes from positive to negative. To do this, we first must find the first derivative:

f'(x)=x^2+16x+64

We found the derivative using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n+1}$

Now, we must find the critical point(s), the point(s) at which the first derivative is equal to zero:

(x-8)^2=0

c=8

Now, we make our intervals over which to analyze the sign of the first derivative:

$(-\infty , 8), (8, \infty )$

Over the first interval, the firt derivative is positive, and over the second interval, the first derivative is positive. Because the first derivative doesn't change from positive to negative, there are no local maxima.

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Example Question #16 : Local Maximum

What is the maximum of y=x^3+3x^2-9x-27 over the interval [-4,2] ?

Possible Answers:

Correct answer:

Explanation:

To find the maximum of a function, find the first derivative. In order to find the derivative of this fuction use the power rule which states, $\frac{d}{dx}x^n=nx^{n-1}$ .

Given the function, f(x)=x^3+3x^2-9x-27 and applying the power rule we find the following derivative.

f'(x)=3x^2+6x-9=3(x^2+2x-3)=3(x+3)(x-1)=0

x=-3,1

Check the -value at each endpoint and when the first derivative is zero, namely x=-4,-3,1,2

f(-4)=-7

f(-3)=0

f(1)=-32

f(2)=-25

The largest value is .

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Example Question #11 : Local Maximum

Find the -value where the local maximum occurs on

$y=\frac{x^2+3x+2}{x-1}$ .

Possible Answers:

$1-\sqrt{6}$

$\frac{7}{2}$

$1+\sqrt{6}$

$-\frac{3}{2}$

Correct answer:

$1-\sqrt{6}$

Explanation:

To find the maximum of a function, find the first derivative. In order to find the derivative of this fuction use the quotient rule which states,

$\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ .

Given the function, $f(x)=\frac{x^2+3x+2}{x-1}$ and applying the quotient rule we find the following derivative.

$f'(x)=\frac{(x-1)(2x+3)-(x^2+3x+2)(1)}{(x-1)^2}=\frac{x^2-2x-5}{(x-1)^2}=0$

x^2-2x-5=0

$x=\frac{2\pm \sqrt{(-2)^2-4(1)(-5)}}{2(1)}=\frac{2 \pm \sqrt{24}}{2}=\frac{2 \pm2 \sqrt{6}}{2} = 1 \pm \sqrt {6}$

f'(x)>0 when $x<1-\sqrt{6}$ and f'(x)<0 when $x>1-\sqrt{6}$ , which indicates that f(x) has a local maximum at $x=1-\sqrt{6}$ .

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Example Question #2501 : Calculus

Find the x-coordinates of all the local maxima of

2x^3+3x^2-12x+5 .

Possible Answers:

-2, 1, 0

None of the other answers.

-2, 1

Correct answer:

Explanation:

We need to differentiate term by term, applying the power rule,

$(x^n)'=nx^{n-1}$

This gives us

$\begin{align*} (2x^3+3x^2-12x+5)' &= 6x^2+6x-12 \end{align*}$

The critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:

$\begin{align*} x &= \frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a} \\ &= \frac{-6\pm\sqrt{36-4\cdot 6\cdot -12}}{12}\\ &= \frac{-6\pm\sqrt{384}}{12}\\ &= \frac{-6\pm18}{12}\\ &=\frac{12}{12} \text{ or } \frac{-24}{12} \\ &=1 \text{ or } -2 \end{align*}$

Any local maximum will fall at a critical point where the derivative passes from positive to negative. To check this, we check a point in each of the intervals defined by the critical points:

$(-\infty, -2), (-2,1) $ and $ (1, \infty)$ .

Let's take -3 from the first interval, 0 from the second interval, and 2 from the third interval.

$6\cdot (-3)^2+6\cdot (-3)-12=30$

$6\cdot 0^2+6\cdot 0-12=-12$

$6\cdot 2^2+6\cdot 2-12=24$

The derivative moves from positive to negative at -2, so that is the function's only local maximum.

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Example Question #341 : Ap Calculus Bc

$\begin{align*}&\text{Determine any local maxima for the function}\\&f(x)=x + 3x^{2} - x^{3} + 1\end{align*}$

Possible Answers:

$-0.155,2.155\text{ = maxima.}$

$2.155\text{ = maxima.}$

$3.383\text{ = maxima.}$

$-0.155\text{ = maxima.}$

Correct answer:

$2.155\text{ = maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }x + 3x^{2} - x^{3} + 1\\&\text{The derivative is }6x - 3x^{2} + 1\\&\text{The roots of this function are }x=-0.155,2.155\\&\text{Checking each, we find that:}\\&\text{The root at }x=-0.155\text{ is a minimum.}\\&\text{The root at }x=2.155\text{ is a maximum.}\\&\\&2.155\text{ = maxima.}\end{align*}$

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Example Question #342 : Ap Calculus Bc

$\begin{align*}&\text{Find any and all local maxima for the function}\\&x + 5x^{2} - 6x^{3} + 2x^{4} - 8x^{5} - 3\end{align*}$

Possible Answers:

$0.5\text{ = maxima.}$

$-0.579\text{ = maxima.}$

$-0.086\text{ = maxima.}$

$0.5,-0.086\text{ = maxima.}$

Correct answer:

$0.5\text{ = maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }x + 5x^{2} - 6x^{3} + 2x^{4} - 8x^{5} - 3\\&\text{The derivative is }10x - 18x^{2} + 8x^{3} - 40x^{4} + 1\\&\text{The roots of this function are }x=0.5,-0.086\\&\text{Checking each, we find that:}\\&\text{The root at }x=0.5\text{ is a maximum.}\\&\text{The root at }x=-0.086\text{ is a minimum.}\\&\\&0.5\text{ = maxima.}\end{align*}$

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Example Question #343 : Ap Calculus Bc

$\begin{align*}&\text{Determine any local maxima for the function}\\&f(x)=5x - 7x^{2} - 8x^{3} - 4\end{align*}$

Possible Answers:

$-0.83333\text{ = maxima.}$

$-0.83333,0.25,-1.5088\text{ = maxima.}$

$-0.83333,0.25\text{ = maxima.}$

$0.25\text{ = maxima.}$

Correct answer:

$0.25\text{ = maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }5x - 7x^{2} - 8x^{3} - 4\\&\text{The derivative is }5 - 24x^{2} - 14x\\&\text{The roots of this function are }x=-0.83333,0.25\\&\text{Checking each, we find that:}\\&\text{The root at }x=-0.83333\text{ is a minimum.}\\&\text{The root at }x=0.25\text{ is a maximum.}\\&\\&0.25\text{ = maxima.}\end{align*}$

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Example Question #344 : Ap Calculus Bc

$\begin{align*}&\text{Find any and all local maxima for the function}\\&7x^{4} - 6x^{2} - 3x^{3} - 3x + 4\end{align*}$

Possible Answers:

$0.917\text{ = maxima.}$

$\text{There are no local maxima.}$

$-0.917\text{ = maxima.}$

$0.641,1.127\text{ = maxima.}$

Correct answer:

$\text{There are no local maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }7x^{4} - 6x^{2} - 3x^{3} - 3x + 4\\&\text{The derivative is }28x^{3} - 9x^{2} - 12x - 3\\&\text{The roots of this function are }x=0.917\\&\text{Checking each, we find that:}\\&\text{The root at }x=0.917\text{ is a minimum.}\\&\text{The root at }x=1.167\text{ is neither a maximum nor minimum.}\\&\\&\text{There are no local maxima.}\end{align*}$

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AP Calculus BC : Derivatives

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Finding Maximums

Example Question #71 : Derivatives

Example Question #1 : Finding Maximums

Example Question #16 : Local Maximum

Example Question #11 : Local Maximum

Example Question #2501 : Calculus

Example Question #341 : Ap Calculus Bc

Example Question #342 : Ap Calculus Bc

Example Question #343 : Ap Calculus Bc

Example Question #344 : Ap Calculus Bc

All AP Calculus BC Resources

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