0
$\begingroup$

I would like to find the intervals on which the function $$\frac{-4x}{x^2-1},$$ increases and decrease, but I am not sure how? Could somebody help me? My textbook it says to first find the intervals and then the critical points. I have no clue how to do this, however.

$\endgroup$
  • $\begingroup$ You should already be acquainted with how to discuss the sign of a fraction, are you not? $\endgroup$ – Saucy O'Path Jun 8 '18 at 10:18
  • $\begingroup$ Do you know the definition of a critical point? $\endgroup$ – Michael Burr Jun 8 '18 at 10:20
0
$\begingroup$

If $f(x)=\frac{-4x}{x^2-1}$ , then $f'(x)=\frac{4+4x^2}{(x^2-1)^2}$

($x \ne 1$).

Hence $f'(x)>0$ for all $x \ne 1$.

Conclusion ?

$\endgroup$
  • $\begingroup$ The equation that i wrote is already deviated... $\endgroup$ – Giulia Della Rosa Jun 8 '18 at 10:28
  • $\begingroup$ It says that f′(x)>0 for x not being equal to -1 and f'(x) smaller than 0 when x is not equal to 1 $\endgroup$ – Giulia Della Rosa Jun 8 '18 at 10:29
0
$\begingroup$

You have to study the sign of the derivative: $$ f'(x)=\frac{-4x}{(x+1)(x-1)} $$ is positive when $x<-1$ or $0\le x < 1$, is negative when $-1< x \le 0$ or $x>1$, and is null when $x=0$.

So there is a local minimum in $x=0$.

$\endgroup$

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.