Question

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

`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

Answer 1

There is no such function.

Explanation:

If `f''(x) = 0` for all `x in RR` then `f` is a linear function of the form `f(x)=mx+b`.
(This is a consequence of the Mean Value Theorem.)

As such, it has derivative `f'(x)=m` so it cannot have a critical number unless `m=0` and `f` is a constant function.

So either `f` has no minimum nor maximum or `f(x) = 0x+b` is a constant function and has minimum = maximum = `b` at every value of `x`.

However no linear function satisfies the given equation
If `f(x) = mx+b` then `f'(x)=m`, so the equation

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

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

`1x^2+0x+m = -mx^2+(5m-b)x+5b`

Setting corresponding coefficients equal we get

`m = -1`, but then we also get

`-5-b=0` so `b=-5`
and
`-1=5b` so `b=-1/5`

`b` cannot have two values so there is no solution to the system.

Answer 2

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

Explanation:

Suppose that for all `x in RR`,
`f''(x) = 0` and
`f'(x)+x^2=(5-x)f(x)`.

Observe that

`f'(x) = (5-x)f(x)-x^2`

Differentiate both sides on the equation, to get

`f''(x) = (-1)f(x)+(5-x)f'(x)-2x`

Use the two facts assumed to get:

`0 = -f(x)+(5-x)underbrace([(5-x)f(x)-x^2])_(f'(x))-2x`

Expand:

`(x^2-10x+24)f(x) +x^3-5x^2-2x = 0`

Solve for `f(x)`

`f(x) = (-x^3+5x^2+2x)/(x^2-10x+24)`

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

`f''(x) = (-48(x^3-15x^2+78x+140))/(x^2-10x+24)^3`

Which is not identically `0`. For example `f''(1) != 0`

So `f` cannot satisfy both conditions given.