Least gradient value for $f(x)=e^{-2x} \tan x$

Question

I’ve derived it into $e^{-2x} (-1+ \tan x)^2$, but i couldn’t think of any other way other than to look at a graph which is π/4. Is there any way to find it through calculation?

Answer 1

Yes your derivation is correct indeed we have

$$f(x)=e^{-2x} \tan x \implies f'(x)=-2e^{-2x} \tan x+e^{-2x} (1+\tan^2 x)=$$$$=e^{-2x}\left(\tan^2 x-2\tan x+1\right)=e^{-2x}\left(\tan x-1\right)^2\ge 0$$

where we have used that

• $e^{-2x}>0$
• $\left(\tan x-1\right)^2\ge 0$

Can you figure out when $f'(x)=0$?

Answer 2

You correctly differentiated to get $f’(x) = e^{-2x}(\tan x-1)^2$. Notice how both factors are non-negative:

$$e^{-2x} > 0$$

$$(\tan x-1)^2 \geq 0$$

Hence, the least value must be $\geq 0$.

By inspection, it’s clear how the second factor can become $0$:

$$\tan x-1 = 0$$

from which you obtain the desired result.