If  has a relative extrema at  and  exists then  is a critical point of .  In fact, it will be a critical point such that .

This is a fairly simple proof.  We’ll assume that  has a relative maximum to do the proof.  The proof for a relative minimum is nearly identical.  So, if we assume that we have a relative maximum at  then we know that  for all x that are sufficiently close to .  In particular for all h that are sufficiently close to zero (positive or negative) we must have,

or, with a little rewrite we must have,

                                                                                                            (1)

Now, at this point assume that  and divide both sides of (1) by h.  This gives,

Because we’re assuming that  we can now take the right-hand limit of both sides of this.

We are also assuming that  exists and recall that if a normal limit exists then it must be equal to both one-sided limits.  We can then say that,

If we put this together we have now shown that .

Okay, now let’s turn things around and assume that  and divide both sides of (1) by h.  This gives,

Remember that because we’re assuming  we’ll need to switch the inequality when we divide by a negative number.  We can now do a similar argument as above to get that,

The difference here is that this time we’re going to be looking at the left-hand limit since we’re assuming that .  This argument shows that .

We’ve now shown that  and .  Then only way both of these can be true at the same time is to have  and this in turn means that  must be a critical point.

As noted above, if we assume that  has a relative minimum then the proof is nearly identical and so isn’t shown here.  The main differences are simply some inequalities need to be switched.

1.      If  for every x on some interval I, then  is increasing on the interval.

2.      If  for every x on some interval I, then  is decreasing on the interval.

3.      If  for every x on some interval I, then  is constant on the interval.

Let  and  be in I and suppose that .  Now, using the Mean Value Theorem on  means there is a number c such that  and,

Because  we know that c must also be in I and so we know that  we also  know that .  So, this means that we have,

Rewriting this gives,

and so, by definition, since  and  were two arbitrary numbers in I,  must be increasing on I.

This proof is nearly identical to the previous part.

Let  and  be in I and suppose that .  Now, using the Mean Value Theorem on  means there is a number c such that  and,

Because  we know that c must also be in I and so we know that  we also  know that .  So, this means that we have,

Rewriting this gives,

and so, by definition, since  and  were two arbitrary numbers in I,  must be decreasing on I.

Again, this proof is nearly identical to the previous two parts, but in this case is actually somewhat easier.

Let  and  be in I.  Now, using the Mean Value Theorem on  there is a number c such that c is between  and  and,

Note that for this part we didn’t need to assume that  and so all we know is that c is between   and   and so, more importantly, c is also in I. and this means that .  So, this means that we have,

Rewriting this gives,

and so, since  and  were two arbitrary numbers in I,  must be constant on I.

Given the function  then,

1. If  for all x in some interval I then  is concave up on I.
   
   
2. If  for all x in some interval I then  is concave down on I.

Let a be any number in the interval I.  The tangent line to  at  is,

To show that  is concave up on I then we need to show that for any x, , in I that,

or in other words, the tangent line is always below the graph of  on I.   Note that we require  because at that point we know that  since we are talking about the tangent line.

Let’s start the proof off by first assuming that .  Using the Mean Value Theorem on  means there is a number c such that  and,

With some rewriting this is,

                                                                                               (2)

Next, let’s use the fact that  for every x on I.  This means that the first derivative, , must be increasing (because its derivative, , is positive).  Now, we know from the Mean Value Theorem that  and so because  is increasing we must have,

                                                                                                                     (3)

Recall as well that we are assuming  and so .  If we now multiply (3) by  (which is positive and so the inequality stays the same) we get,

Next, add  to both sides of this to get,

However, by (2), the right side of this is nothing more than  and so we have,

but this is exactly what we wanted to show.

So, provided  the tangent line is in fact below the graph of .

We now need to assume .  Using the Mean Value Theorem on  means there is a number c such that  and,

If we multiply both sides of this by 1 and then adding  to both sides and we again arise at (2).

Now, from the Mean Value Theorem we know that  and because  for every x on I we know that the derivative is still increasing and so we have,

Let’s now multiply this by , which is now a negative number since .  This gives,

Notice that we had to switch the direction of the inequality since we were multiplying by a negative number.  If we now add  to both sides of this and then substitute (2) into the results we arrive at,

So, again we’ve shown that the tangent line is always below the graph of .

We’ve now shown that if x is any number in I, with  the tangent lines are always below the graph of  on I and so  is concave up on I.

This proof is nearly identical to the proof of 1 and since that proof is fairly long we’re going to just get things started and then leave the rest of it to you to go through.

Let a be any number in I .  To show that  is concave down we need to show that for any x in I, , that the tangent line is always above the graph of  or, 

From this point on the proof is almost identical to the proof of 1 except that you’ll need to use the fact that the derivative in this case is decreasing since .  We’ll leave it to you to fill in the details of this proof.

Suppose that  is a critical point of  such that  and that  is continuous in a region around .  Then,

1. If  then  is a relative maximum.
2. If  then  is a relative minimum.
3. If  then  can be a relative maximum, relative minimum or neither.

First since we are assuming that  is continuous in a region around  then we can assume that in fact   is also true in some open region, say  around , i.e. .

Now let x be any number such that , we’re going to use the Mean Value Theorem on .  However, instead of using it on the function itself we’re going to use it on the first derivative.  So, the Mean Value Theorem tells us that there is a number  such that,

Now, because  we know that  and we also know that  so we then get that,

However, we also assumed that  and so we have,

Or, in other words to the left of  the function is increasing.

Let’s now turn things around and let x be any number such that  and use the Mean Value Theorem on  and the first derivative.  The Mean Value Theorem tells us that there is a number  such that,

Now, because  we know that  and we also know that  so we then get that,

Again use the fact that we also assumed that  to get,

We now know that to the right of  the function is decreasing.

So, to the left of  the function is increasing and to the right of  the function is decreasing so by the first derivative test this means that  must be a relative maximum.

This proof is nearly identical to the proof of 1 and since that proof is somewhat long we’re going to leave the proof to you to do.  In this case the only difference is that now we are going to assume that  and that will give us the opposite signs of the first derivative on either side of  which gives us the conclusion we were after.  We’ll leave it to you to fill in all the details of this.

There isn’t really anything to prove here.  All this statement says is that any of the three cases are possible and to “prove” this all one needs to do is provide an example of each of the three cases.  This was done in The Shape of a Graph, Part II section where this test was presented so we’ll leave it to you to go back to that section to see those graphs to verify that all three possibilities really can happen.

Suppose  is a function that satisfies all of the following.

1.  is continuous on the closed interval [a,b].
2.  is differentiable on the open interval (a,b).
3.  

Then there is a number c such that  and .  Or, in other words  has a critical point in (a,b).

We’ll need to do this with 3 cases.

Case 1 :  on [a,b] where k is a constant.

In this case  for all x in [a,b] and so we can take c to be any number in [a,b].

Case 2 : There is some number d in (a,b) such that .

Because  is continuous on [a,b] by the Extreme Value Theorem we know that  will have a maximum somewhere in [a,b].  Also, because  and  we know that in fact the maximum value will have to occur at some c that is in the open interval (a,b), or .  Because c occurs in the interior of the interval this means that  will actually have a relative maximum at  and by the second hypothesis above we also know that  exists.  Finally, by Fermat’s Theorem we then know that in fact  must be a critical point and because we know that   exists we must have  (as opposed to  not existing…).

Case 3 : There is some number d in (a,b) such that .

This is nearly identical to Case 2 so we won’t put in quite as much detail.  By the Extreme Value Theorem  will have minimum in [a,b] and because   and  we know that the minimum must occur at  where .  Finally, by Fermat’s Theorem we know that .

Suppose  is a function that satisfies both of the following.

1.  is continuous on the closed interval [a,b].
2.  is differentiable on the open interval (a,b).

Then there is a number c such that a < c < b and

Or,

For illustration purposes let’s suppose that the graph of  is,

Note of course that it may not look like this, but we just need a quick sketch to make it easier to see what we’re talking about here.

The first thing that we need is the equation of the secant line that goes through the two points A and B as shown above.  This is,

Let’s now define a new function, , as to be the difference between  and the equation of the secant line or,

Next, let’s notice that because  is the sum of , which is assumed to be continuous on [a,b], and a linear polynomial, which we know to be continuous everywhere, we know that  must also be continuous on [a,b].

Also, we can see that  must be differentiable on (a,b) because it is the sum of , which is assumed to be differentiable on (a,b), and a linear polynomial, which we know to be differentiable. 

We could also have just computed the derivative as follows,

at which point we can see that it exists on (a,b) because we assumed that  exists on (a,b) and the last term is just a constant.

Finally, we have,

In other words,  satisfies the three conditions of Rolle’s Theorem and so we know that there must be a number c such that  and that,