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### Section 4.5 : The Shape of a Graph, Part I

13. For some function, \(f\left( x \right)\), it is known that there is a relative maximum at \(x = 4\). Answer each of the following questions about this function.

- What is the simplest form for the derivative of this function?

**Note :**There really are many possible forms of the derivative so to make the rest of this problem as simple as possible you will want to use the simplest form of the derivative that you can come up with. - Using your answer from (a) determine the most general form of the function.
- Given that \(f\left( 4 \right) = 1\) find a function that will have a relative maximum at \(x = 4\).

**Note :**You should be able to use your answer from (b) to determine an answer to this part.

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Okay, let’s get started with this problem.

The first thing that we’ll do is assume that the derivative exists everywhere. Making assumptions in a math class is generally a bad thing. However, in this case, because we are being asked to come up the form of the derivative all we are really doing here is starting that process. If we can’t find a derivative that will have a relative maximum at the point that also exists everywhere we can come back and change things up. If we can find a derivative that will exist everywhere (which we can as we’ll see) this assumption will help with keeping the derivative as simple as possible.

Now, given that we are assuming that the derivative exists everywhere and we know that if we have a relative maximum at \(x = 4\) then \(x = 4\) must also be a critical point (recall Fermat’s Theorem from a couple of sections ago…). This is another reason for the assumption we made above. Fermat’s theorem requires that the derivative exist at the point in order to know that it is also a critical point.

Next, because we assumed that the derivative exists everywhere (and in particular it exists at \(x = 4\)) we know that in order for it to be a critical point we must also have \(f'\left( 4 \right) = 0\). There are lots and lots of functions that will be zero at \(x = 4\) but probably one of the simplest is use,

\[f'\left( x \right) = x - 4\]This does give \(f'\left( 4 \right) = 0\) as we need, however we have a problem. We can clearly see that if \(x < 4\) we would have \(f'\left( x \right) < 0\) and if \(x > 4\) we would have \(f'\left( x \right) > 0\). This says that \(x = 4\) would have to be a relative minimum and not the relative maximum that we wanted it to be.

Luckily enough for us this is easy to fix. The only problem with our original guess is that the signs of the derivative to the right and left of \(x = 4\) are opposite what we need them to be. Therefore, all we need to do is change them and that can easily be done by multiplying by a negative or,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{f'\left( x \right) = 4 - x}}\]With this choice we still have \(f'\left( 4 \right) = 0\) and now the derivative is positive if \(x < 4\) and negative if \(x > 4\) which means that \(x = 4\) will be a relative maximum.

As noted in the problem statement there are many possible answers to this part. We will be working with the one given above. However, just to make the point here are a sampling of other derivatives all of which come from functions that have a relative maximum at \(x = 4\).

\[f'\left( x \right) = 16 - {x^2}\,\,\,\,\,\,\,\,f'\left( x \right) = 24 + 18x - 6{x^2}\,\,\,\,\,\,\,\,\,f'\left( x \right) = {{\bf{e}}^{4 - \,x}} - 1\,\,\,\,\,\,\,\,\,f'\left( x \right) = \sin \left( {2\pi - \pi x} \right)\]After working the rest of this problem with \(f'\left( x \right) = 4 - x\) you might want to come back and see if you can repeat the problem with one or more of these to see what you get.

Okay, from the previous part we have assumed that the derivative of our function is \(f'\left( x \right) = 4 - x\) and we need to “undo” the differentiation to determine the most general possible function that we could have had.

So, before doing this let’s recall what we know about differentiation. First, let’s recall the following formula,

\[\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]When we differentiate a power of \(x\) the power goes down by one. So, what would we have to differentiate to get \(x\)? We’ll, since the exponent on the \(x\) is 1 we would have had to differentiate an \({x^2}\). However, because we know that that the derivative of \({x^2}\) is 2\(x\) and we want just an \(x\) that would mean that in fact we would have had to differentiate \({\frac{1}{2}}{x^2}\) to get \(x\).

Next, also recall,

\[\frac{d}{{dx}}\left( {kx} \right) = k\]So, using this as a guide is should be pretty simple to see that we would need to differentiate 4\(x\) to get 4.

So, if we put these two parts together it looks like we could use the following function.

\[f\left( x \right) = 4x - {\frac{1}{2}}{x^2}\]The derivative of this function is clearly \(f'\left( x \right) = 4 - x\). However, it is not the most general possible function that gives this derivative.

Do not forget that the derivative of a constant is zero and so we could add any constant onto our function and get the same derivative.

This is one of the biggest mistakes that students make with learning to “undo” differentiation. Any time we undo differentiation there is always the possibility that there was a constant on the original function and so we need to acknowledge that. We usually do this by adding a “ + \(c\) ” onto the end of our function. We use a general \(c\) because we have no way of knowing that the constant would be and this allows for all possible constants.

Therefore, the most general function that we could use to get \(f'\left( x \right) = 4 - x\) is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( x \right) = 4x - {\frac{1}{2}}{x^2} + c}}\]To do this part all we really need to do is plug \(x = 4\) into our answer from the previous part and set the result equal to 1. This will result in an equation with a single unknown value, \(c\). So all we need to do then is solve for \(c\).

Here is the work for this part.

\[1 = f\left( 4 \right) = 4\left( 4 \right) - {\frac{1}{2}}{\left( 4 \right)^2} + c = 8 + c\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,1 = 8 + c\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,c = - 7\]So, it looks like one possible function that will have a relative maximum at \(x = 4\) is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( x \right) = 4x - {\frac{1}{2}}{x^2} - 7}}\]As a final part to this problem, here is a quick graph of this function to verify that it does in fact have a relative maximum at \(x = 4\).