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AP Calculus BC : Finding Maximums

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Example Questions

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 #341 : Ap Calculus Bc

Find the -value where the local maximum occurs on

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

Possible Answers:

$1-\sqrt{6}$

$-\frac{3}{2}$

$\frac{7}{2}$

$1+\sqrt{6}$

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 #11 : Finding Maximums

Find the x-coordinates of all the local maxima of

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

Possible Answers:

-2, 1, 0

-2, 1

None of the other answers.

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 #11 : Derivative As A Function

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

Possible Answers:

$2.155\text{ = maxima.}$

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

$3.383\text{ = maxima.}$

$-0.155\text{ = maxima.}$

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 #11 : Derivative As A Function

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

Possible Answers:

$0.25\text{ = maxima.}$

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

$-0.83333\text{ = maxima.}$

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

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 #12 : Derivative As A Function

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

Possible Answers:

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

$0.917\text{ = maxima.}$

$-0.917\text{ = maxima.}$

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

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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Example Question #345 : Ap Calculus Bc

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

Possible Answers:

$-0.951,1.529,2.845\text{ = maxima.}$

$0.716,2.375\text{ = maxima.}$

$2.375\text{ = maxima.}$

$0.716\text{ = maxima.}$

Correct answer:

$0.716\text{ = maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }8x + 2x^{2} - 4x^{3} - 5x^{4} + 2x^{5} + 8\\&\text{The derivative is }4x - 12x^{2} - 20x^{3} + 10x^{4} + 8\\&\text{The roots of this function are }x=0.716,2.375\\&\text{Checking each, we find that:}\\&\text{The root at }x=0.716\text{ is a maximum.}\\&\text{The root at }x=2.375\text{ is a minimum.}\\&\\&0.716\text{ = maxima.}\end{align*}$

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Example Question #81 : Derivatives

$\begin{align*}&\text{Determine any local maxima for the function}\\&f(x)=5 - 6x^{2} - 3x^{3} - 3x^{4} - 8x^{5} - 3x\end{align*}$

Possible Answers:

$-0.458\text{ = maxima.}$

$-0.346\text{ = maxima.}$

$0.569\text{ = maxima.}$

$-0.458,-0.346\text{ = maxima.}$

Correct answer:

$-0.346\text{ = maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }5 - 6x^{2} - 3x^{3} - 3x^{4} - 8x^{5} - 3x\\&\text{The derivative is }- 12x - 9x^{2} - 12x^{3} - 40x^{4} - 3\\&\text{The roots of this function are }x=-0.458,-0.346\\&\text{Checking each, we find that:}\\&\text{The root at }x=-0.458\text{ is a minimum.}\\&\text{The root at }x=-0.346\text{ is a maximum.}\\&\\&-0.346\text{ = maxima.}\end{align*}$

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Example Question #352 : Ap Calculus Bc

$\begin{align*}&\text{Find the local maxima for }\\&f(x)=2x^{2} - 3x + 4x^{3} - 3x^{4} + 3x^{5} - 2\end{align*}$

Possible Answers:

$-0.481\text{ = maxima.}$

$0.388\text{ = maxima.}$

$0.93\text{ = maxima.}$

$-0.481,0.388\text{ = maxima.}$

Correct answer:

$-0.481\text{ = maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }2x^{2} - 3x + 4x^{3} - 3x^{4} + 3x^{5} - 2\\&\text{The derivative is }4x + 12x^{2} - 12x^{3} + 15x^{4} - 3\\&\text{The roots of this function are }x=-0.481,0.388\\&\text{Checking each, we find that:}\\&\text{The root at }x=-0.481\text{ is a maximum.}\\&\text{The root at }x=0.388\text{ is a minimum.}\\&\\&-0.481\text{ = maxima.}\end{align*}$

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AP Calculus BC : Finding Maximums

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #16 : Local Maximum

Example Question #341 : Ap Calculus Bc

Example Question #11 : Finding Maximums

Example Question #11 : Derivative As A Function

Example Question #342 : Ap Calculus Bc

Example Question #11 : Derivative As A Function

Example Question #12 : Derivative As A Function

Example Question #345 : Ap Calculus Bc

Example Question #81 : Derivatives

Example Question #352 : Ap Calculus Bc

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