Finding Maximums - AP Calculus BC

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Question

Define $f(x) = \frac{x}{2}+ \sin x$ .

Give the maximum value of on the interval $[0, \pi]$ .

Answer

First, we determine if there are any points at which .

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

$f'(x) = \frac{1}{2} + \cos x$

$\frac{1}{2} + \cos x = 0$

$\cos x = -\frac{1}{2}$

The only point on the interval on which this is true is $x =\frac{ 2\pi }{3}$ .

We test this point as well as the two endpoints, and $x = \pi$ , by evaluating for each of these values.

$f(0) = \frac{0}{2}+ \sin 0 = 0 + 0 = 0$

$f\left ( \frac{ 2\pi }{3} \right ) = \frac{ \left ( \frac{ 2\pi }{3} \right )}{2}+ \sin \frac{ 2\pi }{3} = \frac{ \pi }{3} + \frac{\sqrt{3}}{2} \approx 2$

$f( \pi ) = \frac{ \pi}{2}+ \sin \pi = \frac{ \pi}{2} + 0 = \frac{ \pi}{2} \approx 1.6$

Therefore, assumes its maximum on this interval at the point $x =\frac{ 2\pi }{3}$ , and $f\left ( \frac{ 2\pi }{3} \right ) = \frac{ \pi }{3} + \frac{\sqrt{3}}{2} = \frac{ 2\pi + 3\sqrt{3} }{6}$ .

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Question

At which point does a local maxima appear in the following function?

Answer

A local max will occur when the function changes from increasing to decreasing. This means that the derivative of the function will change from positive to negative.

First step is to find the derivative.

Find the critical points (when is or undefined).

$0=3x^2+4x, x=0,-\frac{4}{3}$

Next, find at which of these two values changes from positive to negative. Plug in a value in each of the regions into .

The regions to be tested are $\left(-\infty,-\frac{4}3\right)$ , $\left(-\frac{4}3,0\right)$ , and $(0,\infty)$ .

A value in the first region, such as , gives a positive number, and a value in the second range gives a negative number, meaning that $x=-\frac{4}{3}$ must be the point where the max occurs.

To find what the coordinate of this point, plug in $x=-\frac{4}{3}$ in to , not , to get .

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Question

What is the maximum value of the function on the interval ?

Answer

First, we need to find the critical points of the function by taking .

This is the derivative of a polynomial, so you can operate term by term.

This gives us,

.

Solving for by factoring, we get

.

This gives us critical values of 0 and $-\frac{2}{9}$ . Since we are operating on the interval , we make sure our endpoints are included and exclude critical values outside this interval. Now we know the maximum could either occur at or . As the function is decreasing, we know at the max occurs at and that that value is .

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Question

For the equation , graph the function, and identify where the local minima is.

Answer

By graphing the equation , we can see that there minimum at , and that the graph continues to rise in both directions around this point, so this must be a local minimum. We also know that the graph rises infinitely in both directions, so this must be the only local minimum.

Another way to identify the local minima is by taking the derivative of the function and setting it equal to zero.

Using the power rule,

$f(x)=x^n\rightarrow f'(x)=nx^{n-1}$ we find the derivative to be,

$y=x^2 \rightarrow y'=2\cdot x^{2-1}=2x$ .

From here we set the derivative equal to zero and solve for x. By doing this we will identify the critical values of the function

Now we will plug in the x value and find the corresponding y value in the original equation. We will also plug in an x value that is lower than the critical x value and a x value that is higher than the critical value to confirm whether we have a local minima or maxima.

Since both of the x values have a larger y value than the y value that corresponds to , we know that the minimum occurs at .

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Question

Find the local maximum of the function on the interval .

Answer

To find the local maximum, we must find where the derivative of the function is equal to 0.

Given that the derivative of the function yields using the power rule $y=x^n \rightarrow y'=nx^{n-1}$ . We see the derivative is never zero.

However, we are given a closed interval, and so we must proceed to check the endpoints. By graphing the function, we can see that the endpoint is, in fact, a local maximum.

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Question

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

Answer

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.

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Question

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

Answer

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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Question

What is the maximum of over the interval ?

Answer

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 .

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Question

Find the -value where the local maximum occurs on

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

Answer

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.

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

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

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

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Question

Given that the equation of a graph is find the value of the local maximum on this graph.

Answer

To find the critical points of the graph, you first must take the derivative of the equation of the graph and set it equal to zero. To take the derivative of this equation, we must use the power rule,

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

We also must remember that the derivative of an constant is 0. The derivative of the equation for this graph comes out to $y^{'}=2x+4$ . Solving for when , you find that . The tricky part now is to find out whether or not this point is a local maximum or a local minimum. In order to figure this out we will find whether or not the slope is increasing towards this point or decreasing. Remember that the derivative of a graph equation gives the slope of the graph at any given point.

Thus when we plug in into the slope equation, we find that the slope has a positive value. This means that the slope is increasing as the graph leaves , meaning that this point is a local minimum, We plug in into the slope equation and find that the slope is negative, confirming that is the local minimum. That means that there is no local maximum on this graph.

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Question

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

?

Answer

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.

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Question

Find the local maxima of the following function:

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

Answer

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:

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:

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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Question

Find the x-coordinates of all the local maxima of

.

Answer

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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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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