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AP Calculus BC : Finding Maximums

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Example Question #21 : Derivative As A Function

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

Possible Answers:

$-0.642,0.087\text{ = maxima.}$

$0.087\text{ = maxima.}$

$-0.642\text{ = maxima.}$

$0.713\text{ = maxima.}$

Correct answer:

$-0.642\text{ = maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }5x^{2} - x + 6x^{3} - 4\\&\text{The derivative is }10x + 18x^{2} - 1\\&\text{The roots of this function are }x=-0.642,0.087\\&\text{Checking each, we find that:}\\&\text{The root at }x=-0.642\text{ is a maximum.}\\&\text{The root at }x=0.087\text{ is a minimum.}\\&\\&-0.642\text{ = maxima.}\end{align*}$

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Example Question #82 : Derivatives

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

Possible Answers:

$-1.618,0.618,6.662\text{ = maxima.}$

$-1.069\text{ = maxima.}$

$-1.069,5.305\text{ = maxima.}$

$5.305\text{ = maxima.}$

Correct answer:

$-1.069\text{ = maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }5x^{2} - 8x - 5x^{3} - 6x^{4} + x^{5} + 5\\&\text{The derivative is }10x - 15x^{2} - 24x^{3} + 5x^{4} - 8\\&\text{The roots of this function are }x=-1.069,5.305\\&\text{Checking each, we find that:}\\&\text{The root at }x=-1.069\text{ is a maximum.}\\&\text{The root at }x=5.305\text{ is a minimum.}\\&\\&-1.069\text{ = maxima.}\end{align*}$

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Example Question #21 : Derivative As A Function

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

Possible Answers:

$2.812\text{ = maxima.}$

$1.609\text{ = maxima.}$

$-0.276,1.609\text{ = maxima.}$

$-0.276\text{ = maxima.}$

Correct answer:

$-0.276\text{ = maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }3x^{3} - 6x^{2} - 4x - 8\\&\text{The derivative is }9x^{2} - 12x - 4\\&\text{The roots of this function are }x=-0.276,1.609\\&\text{Checking each, we find that:}\\&\text{The root at }x=-0.276\text{ is a maximum.}\\&\text{The root at }x=1.609\text{ is a minimum.}\\&\\&-0.276\text{ = maxima.}\end{align*}$

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Example Question #81 : Derivatives

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

Possible Answers:

$-0.188\text{ = maxima.}$

$-0.25,-0.188,0.76\text{ = maxima.}$

$-0.879,1.259\text{ = maxima.}$

$-0.25,0.76\text{ = maxima.}$

Correct answer:

$-0.188\text{ = maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }7x^{4} - 4x^{2} - 3x^{3} - x - 4\\&\text{The derivative is }28x^{3} - 9x^{2} - 8x - 1\\&\text{The roots of this function are }x=-0.25,-0.188,0.76\\&\text{Checking each, we find that:}\\&\text{The root at }x=-0.25\text{ is a minimum.}\\&\text{The root at }x=-0.188\text{ is a maximum.}\\&\text{The root at }x=0.76\text{ is a minimum.}\\&\\&-0.188\text{ = maxima.}\end{align*}$

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Example Question #351 : Ap Calculus Bc

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

Possible Answers:

$-0.549\text{ = maxima.}$

$0.322\text{ = maxima.}$

$-1.054\text{ = maxima.}$

$-0.549,0.322\text{ = maxima.}$

Correct answer:

$-0.549\text{ = maxima.}$

Explanation:

$\begin{align*}&\text{Local maxima of a function are relative peaks; points that have}\\&\text{no greater value on either immediate side of them. To find local}\\&\text{maxima, take the derivative of a function and find the roots}\\&\text{of this derivative: the points where it equals zero. If a point}\\&\text{is a local maxima, then derivative values immediately to the}\\&\text{left will be positive, and values to the immediate right will}\\&\text{be negative.}\\&\text{For our function }4x^{2} - 3x + 3x^{3} - 5x^{4} + 3x^{5} + 6\\&\text{The derivative is }8x + 9x^{2} - 20x^{3} + 15x^{4} - 3\\&\text{The roots of this function are }x=-0.549,0.322\\&\text{Checking each, we find that:}\\&\text{The root at }x=-0.549\text{ is a maximum.}\\&\text{The root at }x=0.322\text{ is a minimum.}\\&\\&-0.549\text{ = maxima.}\end{align*}$

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Example Question #22 : Derivative As A Function

Determine the local minima of the following function:

$f(x)=\frac{x^4}{4}+\frac{5x^3}{3}-\frac{4x^3}{6}-5x^2$

Possible Answers:

The function has no local minima

x=-5, 2

x=2

x=-5

x=0

Correct answer:

x=-5, 2

Explanation:

To determine the local maxima of the function, we must determine the points at which the function's first derivative changes from positive to negative.

First, we must find the first derivative of the function:

f'(x)=x^3+5x^2-2x^2-10x

The derivative was found using the following rule:

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

Now, we must find the critical values, for which the first derivative is equal to zero:

x^3+5x^2-2x^2-10x=0

x^2(x+5)+2x(x+5)=0

(x^2-2x)(x+5)=0

c=-5, 0, 2

Using the critical values, we now create intervals over which to evaluate the sign of the first derivative:

$(-\infty, -5), (-5, 0), (0, 2), (2, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, on the second interval, the first derivative is positive, on the third interval, the first derivative is negative, and on the fourth interval, the first derivative is positive. The values at which the first derivative changed from positive to negative were x=-5, 2 , thus a local maximum exists for those two values.

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Example Question #28 : Derivative As A Function

Find the local maxima of the function if its first derivative is given by

f'=(x+4)(x+1)(x+2)

Possible Answers:

x=2

x=-2

x=-4

x=-4, -1

x=-4, -2, -1

Correct answer:

x=-2

Explanation:

To determine the local maxima of the function, we must determine the points at which the function's first derivative changes from positive to negative.

We are given the first derivative of the function; now we must find the critical values, at which the first derivative is equal to zero:

(x+4)(x+1)(x+2)=0

c=-4, -2, -1

Using the critical values, we now create intervals over which to evaluate the sign of the first derivative:

$(-\infty, -4), (-4, -2), (-2, -1), (-1, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, on the second interval, the first derivative is positive, on the third interval, the first derivative is negative, and on the fourth interval, the first derivative is positive. The only value at which the first derivative changed from positive to negative was x=-2 , thus a local maximum exists here.

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Example Question #21 : Derivative As A Function

Determine the maximum value attained by the function

f(x)=2x^3-x^4-1 .

Possible Answers:

$\frac{43}{16}$

$\frac{3}{2}$

$\frac{11}{16}$

Correct answer:

$\frac{11}{16}$

Explanation:

To find the extrema of f(x) , we evaluate the derivative f'(x) and find where it is equal to , keeping in mind that we have to actually test the value of f(x) at these (zero-slope) values of to confirm the function is maximal or minimal there. Therefore we require that

f'(x)=6x^2-4x^3=2x^2(3-2x)=0 .

By the zero product property, this is true when either x=0 or x=3/2 , so any extrema occur at these values of .

Evaluating the function at these values of gives

f(0)=-1

and

$f\left ( \frac{3}{2}\right )=\frac{11}{16}$ ,

but since we are seeking the maximum, we conclude that 11/16 is the maximum value attained by f(x) .

Graph: We see the maximum at x=3/2 as claimed, and a critical point at x=0 , which is neither a local maximum nor minimum.

Screen shot 2016 03 31 at 5.33.31 pm

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Example Question #22 : Derivative As A Function

Compute the maximum value attained by the function

$f(x)=\frac{x}{2x^2+1}$ .

Possible Answers:

$\frac{1}{3}$

$\frac{\sqrt{2}}{3}$

$\frac{\sqrt{2}}{2}$

$\frac{1}{2}$

$\frac{\sqrt{2}}{4}$

Correct answer:

$\frac{\sqrt{2}}{4}$

Explanation:

We use the quotient rule to differentiate f(x) and set the derivative to to find any extrema.

$f'(x)=\frac{(1)(2x^2+1)-(x)(4x)}{(2x^2+1)^2}=0$

1-2x^2=0

(where in the last step we multiplied both sides by (2x^2+1)^2 , which is never )

So any extrema occur at

$x=\pm\frac{1}{\sqrt{2}}$ .

Evaluating the original function at these values of gives:

$f\left (\frac{1}{\sqrt{2}} \right )=\frac{\left (\frac{1}{\sqrt{2}} \right )}{2\left (\frac{1}{\sqrt{2}} \right )^2+1}=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}$

$f\left (-\frac{1}{\sqrt{2}} \right )=\frac{\left (-\frac{1}{\sqrt{2}} \right )}{2\left (-\frac{1}{\sqrt{2}} \right )^2+1}=-\frac{1}{2\sqrt{2}}=-\frac{\sqrt{2}}{4}$ .

We choose the larger value and conclude that the maximum value attained by the function is

$\frac{\sqrt{2}}{4}$ .

Graph: Both extrema can be seen on the graph of the function below.

Screen shot 2016 03 31 at 2.24.47 pm

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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 #21 : Derivative As A Function

Example Question #82 : Derivatives

Example Question #21 : Derivative As A Function

Example Question #81 : Derivatives

Example Question #351 : Ap Calculus Bc

Example Question #22 : Derivative As A Function

Example Question #28 : Derivative As A Function

Example Question #21 : Derivative As A Function

Example Question #22 : Derivative As A Function

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