Finding the highest profits can be done by maximizing .  This is possible by finding it's derivative and setting it equal to zero.  After doing this, we can see that because , we can now solve for our answer,.

We take the derivative:

To find the critical points, we set  

Solving for x using the quadratic formula:

Which is our answer

Take the derivative of the function:

To find critical points, we set dy/dx = 0

or, equivalently,

The first couple points at which this is true are

.

These are both local maxima, but only  fulfils the condition

. Hence,  is our answer.

To find the critical points, we allow .

We notice right away that  is a critical point but it is not between 0 and 2.

we also notice that  is a critical point also. It is, indeed, between 0 and 2. The next possible critical point is  which is greater than 2. Evaluating the function at the two points suggests that the function is a minimum at  . Hence, it is our answer.

To find local extrema (if any), set the derivative equal to 0 and solve for .

Here, , so .

However, this function is a parabola (since it is quadratic) and concave up (since the coefficient of the quadratic term is positive), and therefore has no local maximum -- only a local minimum.

To find local extrema, use the first derivative test.

Here, , so  or .

Test points in each of the intervals defined by those potential extrema. For example,  (increasing),  (decreasing), and  (increasing).

Since the function increases up to , then decreases, is a local maximum. (For completeness,  is a local minimum, not a maximum.)

To find a local maximum of a function,  we need to find where the slope of the function changes from positive to negative. This is a point at which the output (the y-value in this case) has increased as much as it can (in the local area of the graph), and just after which the value will decrease. 

The slope of this point must be zero, since, if the function is continuous, the slope must be zero when shifting from a positive value to a negative value. 

To find where the slope of the function is zero, we will find the derivative of the function, and set it equal to zero:

To solve for the x-values that make this statement true (in other words, to find the x-values at which the slope equals zero), we can use the quadratic formula:

having a slope of zero simply means that a value is eligible to be a local maximum or minimum, however. To determine which value is the maximum and which is the minimum, we will plug in numbers into y' on either side of -3*.

So, before x=-3, the slope of the function is positive, and after, the slope is negative. This means that the slope changes from postivie to negative at x=-3. Therefore, The local maximum of the function is found at x=-3.

*Note that we could just as easily plug in numbers on either side of -1/3. Doing this would simply show that the slope is positive before x=-1/3 and positive after it, meaning that there is a local minimum at x=-1/3.

To find the local max of this function, you have to first find the derivative so you cna test values and see what's going with the slope. To find the derivative, take your exponent and multiply it by the leading coefficient and then reduce the exponent by 1. Therefore, your derivative function is: . Then, set that equal to 0 to get your critical points. . Then, set up a number line and test the regions between each critical point. When you pick a test value, plug it into the derivative function. To the left of , the function has a positive slope. Then, after , it becomes negative. After 0, it turns positive again. A local max indicates that the slope went from positive to negative. This is occuring at .

To determine the local maxima of the function, we must find the x-value(s) at which the first derivative of the function changes from positive to negative. 

First, we must find the first derivative:

The derivative was found using the following rule:

Next, we must find the critical values, the values at which the first derivative is equal to zero:

Now, we must use the critical values to create the intervals on which we analyze whether the first derivative is positive or negative:

Note that at the bounds of the intervals the derivative is neither positive nor negative.

To check the sign of the first derivative, simply plug in any point on each interval into the first derivative function: on the first interval, the first derivative is positive, on the second it is negative, and on the third it is positive. The change from positive to negative happened from the first to second interval, so  is the local maximum.

We need to differentiate term by term, applying the power rule,

This gives us

The critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula: