show that $|f|$ has a maximum at a point $x$ satisfying $f^{\prime}(x)=0$, $x=a$, or $x=b$.

Question

let f(x) be a continuous and differentiable function on the interval $[a,b]$ to $\mathbb{R}$.

• show that $|f|$ has a maximum at a point $x$ satisfying $f^{\prime}(x)=0$, $x=a$, or $x=b$.

My proof:

note that :

• If $f$ is continuous in the closed and bounded interval [a,b], then $f$ must attain a maximum and a minimum
• If $f(x)$ is a continuous function on a closed interval $[a,b]$, then $f$ always has an absolute maximum and an absolute minimum value on $[a, b]$.

Reference :

Answer 1

I think what they meant to say is "show that $|f|$ has a maximum at a point $x$ satisfying $f^{\prime}(x)=0$, $x=a$, or $x=b$."

Hint: As you pointed out, a maximum of $|f|$ is attained on $[a,b]$. If that maximum is on the boundary, we are done. Otherwise, let $x\in(a,b)$ be that maximum. Note that $x$ is a maximum or minimum of $f$ (why?), from which the result follows.

Answer 2

Hint. Assume $f$ has a maximum value within $(a..b)$, say at $x ∈ (a..b)$. Now look at the difference quotients $\frac{f(x) - f(x+h)}{h}$ and compare them with zero for $h > 0$ and $h < 0$ respectively. Take limits.

Now, a maximum value of $\lvert f \rvert$ is a maximum value of either $f$ or $-f$. From the above you can characterize the points where they occur.

Answer 3

$f$ being continuous and differentiable, it achieves its absolute extrema where $f'(x)=0$ or at $x=a$ or at $x=b$.

Then the absolute maximum of $|f|$ is the largest of the absolute maximum of $f$ and the absolute maximum of $-f$, which is minus the absolute minimum of $f$. As said, these occurs at $f'(x)=0$ or $x=a$ or $x=b$.

Answer 4

The function $|f|$ is continuous on $[a,b]$, hence it takes its maximum at some point $p\in[a,b]$.

If $p$ is an interior point of this interval and $|f(p)|=0$ then $f(x)\equiv0$, a trivial case. Otherwise assume $f(p)>0$. Then for all $|h|\ll1$ one has $f(p+h)>0$, and therefore $$|f(p+h)|-|f(p)|=f(p+h)-f(p)=\bigl(f'(p)+o(1)\bigr)h\qquad(h\to0)\ .$$ If $f'(p)\ne0$ then the right hand side is positive for suitable $h$ near $0$, hence $p$ cannot be the point where $|f|$ attains its max on $[a,b]$. It follows that necessarily $f'(p)=0$ if $p\in\>]a,b[\>$.

The conclusion is that $${\rm argmax}_{\,a\leq x\leq b}|f(x)|\subset\{a,b\}\cup C\ ,$$ where $C$ denotes the set of critical points of $f$ in $\>]a,b[\>$ .