# Thread: what are maxima, minima and inflection pts

1. ## what are maxima, minima and inflection pts

given the function the function:

$f(x) = xsinx$

we must find its maxima, minima and inflection pts and show that all these pts belong to four distinct curves of equation:

$\frac{x^2}{\sqrt{1+x^2}}$; $-\frac{x^2}{\sqrt{1+x^2}}$; $\frac{2x}{\sqrt{4+x^2}}$; $-\frac{2x}{\sqrt{4+x^2}}$

the maxima minima values are kinda easy. its the inflection points that are, first of all hard to find, and then prove that they belong to two of the curves (the last two).
can anybody come up with some hints to point me in the right direction.
thanks.

2. Hello,
Originally Posted by tombrownington
its the inflection points that are, first of all hard to find, and then prove that they belong to two of the curves (the last two).
can anybody come up with some hints to point me in the right direction.
We have $f(x) = x\sin x$ so $f'(x) = \sin x + x\cos x$ and $f''(x) = 2\cos x-x\sin x$. The point $(x,f(x))$ is an inflection point iff $f''(x)=0\Longleftrightarrow 2\cos x = x\sin x \,\,\,\,\,\,(1)$. As $x=0$ and $x=\frac{\pi}{2}+k\pi,\,k\in\mathbb{Z}$ aren't solutions of $f''(x)=0$, we can safely divide (1) by $x\cos x$ to get $f''(x)=0\Longleftrightarrow \tan x =\frac{2}{x}$. Now we want to show that if $\tan x_0 =\frac{2}{x_0} \,\,\,\,\,\,(2)$ then $(x_0,f(x_0))$ lies on the curve $y=\pm \frac{2x}{\sqrt{4+x^2}}$ that is to say that $x_0\sin x_0 =\pm \frac{2x_0}{\sqrt{4+x^2_0}}$. If you manage to find an expression of $\sin x$ in terms of $\tan x$ and then substitute (2) in this expression, you should get the desired result.

3. thanks flyingsquirrel, its definitely a big step forward.
however i'm not clear about one of your definitions. how is (x0,f(x0)) an inflection point iff f''(x)=0 for x=x0?
shouldn't it be f'(x0)=f''(x0)=0?
thanks anyway for the help

4. Originally Posted by tombrownington
however i'm not clear about one of your definitions. how is (x0,f(x0)) an inflection point iff f''(x)=0 for x=x0?
shouldn't it be f'(x0)=f''(x0)=0?
You're right, I can't use "iff" in this case. However, with you definition, $(0,0)$ is not an inflection point for $x\mapsto \sin x$ which is strange, isn't it ? The origin is an inflection point of the curve of the sine function.

Anyway, in my previous post I only used the implication "if $x_0$ is an inflection point then $f''(x_0)=0$" which is correct.