0
$\begingroup$

!http://puu.sh/ywbUV/42e664f5c7.png

Above is a link to the question! Please correct me if I'm wrong but, I'm guessing that the tangents lines at +-1 of the graph do not exist because the tangent is a vertical line which is undefined.

I feel like i'm supposed to solve this quesiton using Mean Value Theorem...

$\endgroup$
  • $\begingroup$ Yes, $\tan(x)=\pm1$ is undefined because the tangent would just be a horizontal line that never intersects the x-axis. $\endgroup$ – Badr B Nov 29 '17 at 23:52
0
$\begingroup$

You hit it right on the head. As $|x|$ approaches $1$, the tangent line's slope approaches infinity because of the zero in the denominator.

$\endgroup$
0
$\begingroup$

Whenever a function has zero slope, its inverse has an infinite slope so its derivative is not defined.

A simple example is $f(x) = x^2$ on $[0, 1]$. Its slope at $0$ is $0$, so the inverse function, $\sqrt{x}$, has infinite slope there.

$\endgroup$
  • $\begingroup$ How is there an infinite slope from [0,1] in $\sqrt{x}$? It doesn't go to infinity if it's a closed interval rgiht?... $\endgroup$ – user13123 Nov 30 '17 at 5:12
  • $\begingroup$ The slope at 0 goes to $\infty$. $(\sqrt{x})' = (x^{1/2})' = (1/2)x^{-1/2}$. $\endgroup$ – marty cohen Nov 30 '17 at 7:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.