# Show that #f# has no extremas? (Probably using Fermat's Theorem)

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#f# is 2 times differentiable in #RR# with #f''(x)=0# , #x# #in# #RR#

& #f'(x)+x^2=(5-x)f(x)#

Show that #f# has no extremas?

Tip: Use Fermat theorem

&

Show that

Tip: Use Fermat theorem

##### 2 Answers

There is no such function.

#### Explanation:

If

(This is a consequence of the Mean Value Theorem.)

As such, it has derivative

So either

**However no linear function satisfies the given equation**

If

#f'(x)+x^2 = (5-x)f(x)# becomes:

#m+x^2 = -(x-5)(mx+b)# which requires

Setting corresponding coefficients equal we get

and

To see that there is no such function without implicitly using the Mean Value Theorem, please see below.

#### Explanation:

Suppose that for all

Observe that

Differentiate both sides on the equation, to get

Use the two facts assumed to get:

Expand:

Solve for

But then, with a bit of work, (or electronic help)

Which is **not** identically

So