To determine the maximum height of the ball, the first derivative of the function must be found and set equal to zero.

Test values to the left and right to confirm if this point is a maximum.

Because the derivative goes from positive to negative, we know the original function goes from increasing to decreasing and thus this point is a maximum.

Finally, to find the height of the ball at this point, plug in  into the original function.

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:

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. 

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

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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.  

In order to solve this equation, we must first underestand that by taking the derivative of an equation of a graph and setting it equal to zero, we can find the values of  where there are critical points.  Critical points are either local maxima or minima, in order to figure out which you simply plug in numbers before and after that value of  in order to see whether or not the slope increases/decreases as is approaches or leaves that  value.  

In order to take the derivative of equation, the power rule must be applied, .  

Taking the derivative, we find that the equation becomes . 

Setting the equation equal to zero and solving for , we find that the critical points of this graph are .  In order to determine whether or not these points are maxima or minima, we plug in numbers larger/smaller as well as inbetween the two critical points to find the behavior of the slope.  

For this problem I will choose the numbers -5,0 and 1.  These numbers are choosen since   is approximately -4 and   is approximately 0.25.  By choosing numbers that are larger and smaller then the critical points as well as a number inbetween the two critical points, we can see the behavior of the slope throughout the entire graph.

Plugging in those numbers into , we find that  makes  positive,  makes  negative, and  makes  positive.  This means that the slope is positive approaching  and negative leaving it.  The slope is negative approaching  and positive leaving it.  For a critical point to be the local maxima, its slope must be positive approaching it and negative leaving it, therefore  is the local maxima in this graph.

 and we need to find maxima. So lets take the first derivative by the product rule where . In this case, we can set . Next, we can find the derivatives of each . Now we assemble by the product rule above to get:

.

Now we must find critical points (where , endpoints, or undefined values). 

The domain of the function is all real numbers and the endpoints are  and . So now we just need to find where  . This will be equal to  when  and where . There is no value of  we can choose such that  so lets focus on . This is true where . We know that , so maybe we can use this to find values for which . We know that in the second quadrant, sine is positive and cosine is negative. We can check and verify that . The next value will occur in the 4th quadrant, when sine and cosine flip signs. So we can characterize these values that satisfy the equation  as  where . 

Now we need to determine which of these critical points are maxima.  so, the function is increasing from  and we can show this is not a maxima. Since we're going to be testing a lot of points, let's take the second derivative and use the second derivative test. If , we have a max (think concave up). So . 

At , , as  (is in the second quadrant, where only sine is positive).  Therefore,  is a local maximum. The next value, , is in the fourth quadrant where cosine is positive, so the second derivative should be positive, meaning it should be a local minimum. The next point,  should be another maximum, but it lies outside the domain. So the last endpoint, , is the last local maximum on the interval.

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:

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:

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. 

To find the local extrema of a function you must find the zeros of the derivative. The derivative of  is , so we can calculate the derivative of our polynomial and pull out a constant, getting:

We can use the quadratic formula or just guess some easy values, and figure out that the zeros of this polynomial are x=6 and x=2. 2 is the only one of these in the interval stated, so we need to check and see if it's a maximum or a minimum. We have to plug at least three values into our function to see this, so we'll choose x=1, 2, 3.

We can see that as x approaches 2, f(x) is increasing, and as it moves away from 2, f(x) decreases.

So  must be our local maximum.

To find the maximum, you must first find the derivative of the function, which is  We found this derivative using the power rule which states,

.

Then, you need to set that equal to zero to get your critical points.

It will be helpful to factor it:

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Then, set each expression to 0 to get those points:

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Then, put these points on a number line to then test the sign of the slope in between each point.

To the left of , plug in a value (say, 0) into the derivative. You get a positive value. In between and 1, the value is negative. To the right of 1, the value is positive. Then, to find the max, look for where the signs change from positive to negative.

That is occurring at .

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, .

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 .

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, 

.

Given the function,  and applying the quotient rule we find the following derivative.

when and  when , which indicates that has a local maximum at .