Find the absolute maximum of $\frac{x+\sin(x)}{x}$

Question

This is so far what I did.

$f\prime(x) = \frac{cos(x).x-sin(x)}{x^2}$

$\frac{cos(x).x-sin(x)}{x^2} =0$

$cos(x).x = sin(x)$

$x=tan(x)$

I have no clue what to do from here.

This is the graph of y = x and y = tan(x)

Answer 1

Since $\frac{\sin(x)}{x} \leq 1$ and $\lim_{x \to 0} \frac{\sin(x)}{x}=1$ you ahve

$$f(x) \leq 2 \mbox{and} \lim_{x \to 0} f(x)=2$$

So, if $f(0)$ is defined by continuity, the max is 2, otherwise the function has no absolute max.

Answer 2

Looks like you just got off to a bad start: $$f'(x)={(1+\cos x)x-(x+\sin x)\cdot 1\over x^2}={x\cos x-\sin x\over x^2},$$ so $f'(x)=0$ when $x\cos x-\sin x=0\implies x=\tan x$.

Now we need to know over what interval you want an absolute maximum. (Perhaps you want to know if there is global maximum?)

Anyway, here is the graph of $f(x)$, which provides much insight.

Answer 3

$x=0$ should do it. In fact it is the only solution. You can prove $$\tan(x)>x$$ for $x\in(0,\frac{\pi}{2})$ by showing $$ \frac{\tan(x)}{x}>1 $$ Using $$\frac{\tan(x)-0}{x-0}= \tan'(\xi)=\frac{1}{\cos^2(\xi)}>1$$ you see it is true.