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

## $f$ is 2 times differentiable in $\mathbb{R}$ with $f ' ' \left(x\right) = 0$, $x$$\in$$\mathbb{R}$ & $f ' \left(x\right) + {x}^{2} = \left(5 - x\right) f \left(x\right)$ Show that $f$ has no extremas? Tip: Use Fermat theorem

Jan 5, 2018

There is no such function.

#### Explanation:

If $f ' ' \left(x\right) = 0$ for all $x \in \mathbb{R}$ then $f$ is a linear function of the form $f \left(x\right) = m x + b$.
(This is a consequence of the Mean Value Theorem.)

As such, it has derivative $f ' \left(x\right) = m$ 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 $f \left(x\right) = 0 x + b$ is a constant function and has minimum = maximum = $b$ at every value of $x$.

However no linear function satisfies the given equation
If $f \left(x\right) = m x + b$ then $f ' \left(x\right) = m$, so the equation

$f ' \left(x\right) + {x}^{2} = \left(5 - x\right) f \left(x\right)$ becomes:

$m + {x}^{2} = - \left(x - 5\right) \left(m x + b\right)$ which requires

$1 {x}^{2} + 0 x + m = - m {x}^{2} + \left(5 m - b\right) x + 5 b$

Setting corresponding coefficients equal we get

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

$- 5 - b = 0$ so $b = - 5$
and
$- 1 = 5 b$ so $b = - \frac{1}{5}$

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

Jan 5, 2018

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

#### Explanation:

Suppose that for all $x \in \mathbb{R}$,
$f ' ' \left(x\right) = 0$ and
$f ' \left(x\right) + {x}^{2} = \left(5 - x\right) f \left(x\right)$.

Observe that

$f ' \left(x\right) = \left(5 - x\right) f \left(x\right) - {x}^{2}$

Differentiate both sides on the equation, to get

$f ' ' \left(x\right) = \left(- 1\right) f \left(x\right) + \left(5 - x\right) f ' \left(x\right) - 2 x$

Use the two facts assumed to get:

$0 = - f \left(x\right) + \left(5 - x\right) {\underbrace{\left[\left(5 - x\right) f \left(x\right) - {x}^{2}\right]}}_{f ' \left(x\right)} - 2 x$

Expand:

$\left({x}^{2} - 10 x + 24\right) f \left(x\right) + {x}^{3} - 5 {x}^{2} - 2 x = 0$

Solve for $f \left(x\right)$

$f \left(x\right) = \frac{- {x}^{3} + 5 {x}^{2} + 2 x}{{x}^{2} - 10 x + 24}$

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

$f ' ' \left(x\right) = \frac{- 48 \left({x}^{3} - 15 {x}^{2} + 78 x + 140\right)}{{x}^{2} - 10 x + 24} ^ 3$

Which is not identically $0$. For example $f ' ' \left(1\right) \ne 0$

So $f$ cannot satisfy both conditions given.