how to do make the inverse of this function, I already have the answer but I cant see how it got there:
this is the answer:
$$\pm \sqrt{-\tan^{-1}(1-x)}$$
reverse of $f(x) = 1 + \tan(x^2)$
1
$\begingroup$
inverse
radicals
chain-rule
tangent-line
2 Answers
3
Hint
Isolate $y$
$$x=1+\tan(y^2)$$
Also remember that $\tan x$ is an odd function. It means that $-\tan^{-1}(1-x)=\tan^{-1}(x-1)$
Can you finish?
2
Let $y=f(x)=1+\tan(x^2)$
$\implies\tan(x^2)=y-1\implies x^2=\arctan(y-1)\implies x=\pm\sqrt{\arctan(y-1)}$
$\implies f^{-1}(y)=x=\pm\sqrt{\arctan(y-1)}$