There are 2 steps to solving this problem.

First, take the derivative of 

Then, replace the value of x with the given point and evaluate

For example, if  , then we are looking for the value of  , or the derivative of  at .

Calculate 

Derivative rules that will be needed here:

• Derivative of a constant is 0. For example, 
• Taking a derivative on a term, or using the power rule, can be done by doing the following:

Then, plug in the value of x and evaluate

The equation for the derivative at a point is given by

.

By substituting , , we obtain

To find the heat flux, we must find the change in temperature with respect to x:

The derivative was found using the following rule:

Plugging this into Fourier's law, we get 

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

The only point on the interval on which this is true is .

We test this point as well as the two endpoints,  and , by evaluating  for each of these values.

Therefore,  assumes its maximum on this interval at the point , and .

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,

 we find the derivative to be,

.

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 .

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).

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 ,, and .

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  must be the point where the max occurs.

To find what the  coordinate of this point, plug in  in to , not , to get .

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 . 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.

 First rewrite  :

Use the multiplication rule to take the derivative:

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

...and solve for ...

   *

* 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 . 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:

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

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 . 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 .

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,  

.  

We also must remember that the derivative of an constant is 0.  The derivative of the equation for this graph comes out to .  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.