# What are the points of inflection, if any, of f(x)=secxtanx  on [0,2pi]?

Jul 14, 2016

There aren't any as far as I can tell. Once I conclude that any critical points are complex you're probably safe to bail out, tbh I only did the rest out of my own interest in the problem.

#### Explanation:

Ugh, why is it never nice, easy functions...

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\sec x\right) \tan x + \sec x \frac{d}{\mathrm{dx}} \left(\tan x\right)$

$f ' \left(x\right) = \sec x {\tan}^{2} x + {\sec}^{3} x$

$f ' ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\sec x\right) {\tan}^{2} x + \sec x \frac{d}{\mathrm{dx}} \left({\tan}^{2} x\right) + \frac{d}{\mathrm{dx}} \left({\sec}^{3} x\right)$

$f ' ' \left(x\right) = \sec x {\tan}^{3} x + 2 {\sec}^{3} x \tan x + 3 {\sec}^{3} x \tan x = \sec x {\tan}^{3} x + 5 {\sec}^{3} \tan x$

We now have all the derivatives we're going to need. First, we identify the critical points of the function by setting the first derivative equal to zero:

$\sec x {\tan}^{2} x + {\sec}^{3} x = 0$

$\sec x \left({\tan}^{2} x + {\sec}^{2} x\right) = 0$

Recall that ${\tan}^{2} x = {\sec}^{2} x - 1$

(if you want to derive this identity, try dividing every term in ${\sin}^{2} x + {\cos}^{2} x = 1$ by ${\cos}^{2} x$)

$\sec x \left(2 {\sec}^{2} x - 1\right) = 0$

Hence either

$\sec x = 0$ or $2 {\sec}^{2} x - 1 = 0$

Secant is never zero, so no solutions from the first expression.

${\sec}^{2} x = \frac{1}{2}$

$\sec x = \pm \frac{1}{\sqrt{2}}$

Taking the reciprocal:

$\cos x = \pm \sqrt{2}$

$x = {\cos}^{- 1} \left(\sqrt{2}\right)$ remembering that cosine is an even function so $\pm \sqrt{2}$ will yield the same answer.

So, this isn't very nice and is about to get complex. I'm pretty sure this means that there isn't any real points of inflection but I'm going to soldier bravely into it anyway, mostly out of self-interest but you might be interested too? :)

$\textcolor{red}{\text{DISCLAIMER}}$

$\rightarrow$ Depending on how much work you've done with complex numbers, this might just be an unintelligible mess (it might be anyway!). If you haven't really met them yet, i think you can be pretty safe in abandoning ship here.

We let x be some complex number a + bi

$a + b i = {\cos}^{- 1} \left(\sqrt{2}\right)$

$\cos \left(a + b i\right) = \sqrt{2}$

Using Euler's formula, we can see that:

$\cos x = \frac{{e}^{i x} + {e}^{- i x}}{2}$

So we have:

$\frac{{e}^{i \left(a + b i\right)} + {e}^{- i \left(a + b i\right)}}{2} = \sqrt{2}$

${e}^{a i - b} + {e}^{- a i + b} = 2 \sqrt{2}$

${e}^{a i} {e}^{- b} + {e}^{- a i} {e}^{b} = 2 \sqrt{2}$

Using Euler's formula:

${e}^{- b} \left[\cos \left(a\right) + i \sin \left(a\right)\right] + {e}^{b} \left[\cos \left(a\right) - i \sin \left(a\right)\right] = 2 \sqrt{2}$

Collecting real and imaginary parts

$\left({e}^{b} + {e}^{- b}\right) \cos \left(a\right) + i \left({e}^{- b} - {e}^{b}\right) \sin \left(a\right) = 2 \sqrt{2}$

So

$\left({e}^{b} + {e}^{- b}\right) \cos \left(a\right) = 2 \sqrt{2}$

and

$\left({e}^{- b} - {e}^{b}\right) \sin \left(a\right) = 0$

Looking at our second equation, there are a number of solutions that can arise from this. We might think that b = 0 would work, but subbing into the first gives:

$\cos \left(a\right) = \sqrt{2}$

which is the exact same thing that got us into this mess so not helpful. If we try a = 0 we get

${e}^{b} + {e}^{- b} = 2 \sqrt{2}$

This is more promising, try substitution of $u = {e}^{b}$

$u + \frac{1}{u} = 2 \sqrt{2}$

${u}^{2} - 2 \sqrt{2} u + 1 = 0$

$u = \frac{2 \sqrt{2} \pm 2}{2} = \sqrt{2} \pm 1 = {e}^{b}$

$b = \ln \left(\sqrt{2} \pm 1\right)$

Hence the roots are given by

$x = 2 n \pi + \ln \left(\sqrt{2} \pm 1\right) i$

When subbed into second derivative I got $\pm \frac{9}{4} i$ so no inflection points.