Finding Maximums

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AP Calculus BC › Finding Maximums

Questions 1 - 10

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

$0.716\text{ = maxima.}$

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

$2.375\text{ = maxima.}$

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

Explanation

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

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

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

$-0.917\text{ = maxima.}$

$0.917\text{ = maxima.}$

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

Explanation

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

What is the maximum of over the interval ?

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, and applying the power rule we find the following derivative.

Check the -value at each endpoint and when the first derivative is zero, namely

The largest value is .

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

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

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

$\pm3=x$

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

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

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

$-0.642\text{ = maxima.}$

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

$0.087\text{ = maxima.}$

$0.713\text{ = maxima.}$

Explanation

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

Determine the maximum value attained by the function

$\frac{11}{16}$

$\frac{3}{2}$

$\frac{43}{16}$

Explanation

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

By the zero product property, this is true when either or , so any extrema occur at these values of .

Evaluating the function at these values of gives

and

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

but since we are seeking the maximum, we conclude that is the maximum value attained by .

Graph: We see the maximum at as claimed, and a critical point at , which is neither a local maximum nor minimum.

Screen shot 2016 03 31 at 5.33.31 pm

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

$2.155\text{ = maxima.}$

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

$-0.155\text{ = maxima.}$

$3.383\text{ = maxima.}$

Explanation

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

Find the local maximum of the curve $y=\frac{x^5}{e^x}$ .

and

$x=\pm 5$

Explanation

First rewrite $y=\frac{x^5}{e^x}$ :

$y=x^5e^{-x}$

Use the multiplication rule to take the derivative:

$\frac{dy}{dx}=(5x^4)(e^{-x})+(x^5)(-e^{-x})$

$\frac{dy}{dx}=5x^4e^{-x}-x^5e^{-x}$

To find the local extrema, set this to 0...

$0=5x^4e^{-x}-x^5e^{-x}$

...and solve for ...

$5x^4e^{-x} = x^5e^{-x}$ *

$5e^{-x} = xe^{-x}$

* Since we divided by , we have to remember that is a valid solution

Therefore, we know that we have two potential local extrema: and .

By plugging these in, we get two potential local extrema: and $\left(5,\frac{3125}{e^5}\right)$ . Therefore, we know that the slope is positive between and . This means that can't be a local maximum, leaving only as a potential answer.

Next, we can find the slope at . It is:

$\frac{dy}{dx}=5x^4e^{-x}-x^5e^{-x}=5\cdot6^4e^{-6}-6^5e^{-6}=-1296e^-6$

This is negative, meaning that we go from a positive slope to a negative slope at , making it a local maximum.

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

$-0.481\text{ = maxima.}$

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

$0.388\text{ = maxima.}$

$0.93\text{ = maxima.}$

Explanation

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

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

$-0.188\text{ = maxima.}$

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

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

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

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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