Vtyjkyuk

Suppose $a, b in mathbbR$ with $a < b$. Suppose $f, f^prime, f^prime prime$ exist on $(a,b)$. Prove that if $f$ has a local maximum at $c in (a,b)$, then $f^primeprime(c) leq 0$.

I can see why this is true because if $f^primeprime(c) geq 0$, the function $f$ would be concave up and a point to the right of $c$ would be a local maximum, but I'm struggling to write a rigorous proof.

I know that since $c$ is a local max of $f$, we know that $f^prime(c) = 0$. Then
$f^prime(c) = lim_xto c fracf(x) - f(c)x - c = 0$. I also know that $f^primeprime(c) = lim_x to c fracf^prime(x) - f^prime(c)x - c = lim_x to cfracf^prime(x)x - c$ but I don't know how to continue.

I assume $f"$ is continous. Suppose $f"(c)> 0$. Since $f'(c) =0$, There exists an interval $I$ such that $f'(x) >f'(c) =0$ for $x$ in $I, x>c$.This implies that $f$ strictly increases on $[c, x]. $ Contradiction.

We can show this using Taylor`s expansion aswell. We can take the first order taylor expansion around $x_0$ with Lagrange's remainder :
$$f(x)=f(x_0)+f'(x_0)(x-x_0)+frac f''(c_x)(x-x_0)^2 2$$ for some $c_xin(x_0,x)$ or $(x,x_0) $ and $x_0in(a,b)$.

Since $c$ is a local maximum, we have $f'(c)=0$:

$$f(x)=f(c)+frac f''(c_x)(x-x_0)^2 2$$
Since $frac (x-x_0)^2 2geq 0$, The sign of $f''(c_x)$ determines if we have a maximum or a minimum.