Posted by **Leasha** on Tuesday, January 1, 2013 at 12:38pm.

Find the extreme value(s) of the function f(x)=1/(1-x^2).

I tried, and for the first derivative I got -1/(2*sqrt((1-x^2)^3)). I then found critical values of the function to be -1 and 1.

However, the textbook says that the minimum value is 1 at f(0), and I have absolutely no idea where they came up with that answer... Help please!

- Calculus -
**David Q/R**, Tuesday, January 1, 2013 at 3:47pm
I think there's posibly a piece of information missing in the question as you've stated it, which is that you're only supposed to be looking at the function between the values X = -1 and X = +1, because the function is discontinuous at both of those values, and also takes smaller values than f(X) = 1 outside that range: try X = -2 or +2, and you'll get f(X) = -1/3, which is certainly less than 1.

Secondly, I think you've got the differentiation wrong: I get df/dx = 2x/(1-x^2)^2, which if you set it to zero gives you just one real turning point at x = 0, for which f(x) = 1 as stated in the textbook.

## Answer this Question

## Related Questions

- Please check my Calculus - 1. Find all critical values for f(x)=(9-x^2)^⅗ ...
- Please check my Calculus - 1. Which of the following describes the behavior of f...
- Please check my Calculus - 1. Find all points of inflection of the function f(x...
- Please check my Calculus - 1. Find all points of inflection of the function f(x...
- Calculus - another question - "Find the absolute minimum value of the function f...
- Calculus - For y=(1/4)x^4-(2/3)x^3+(1/2)x^2-3, find the exact intervals on which...
- Calculus - Given the function: f(x) = x^2 + 1 / x^2 - 9 a)find y and x ...
- Calculus - For f(x)=2(x+5)^3 +7 Find and classify the extreme values, determine ...
- Calculus - Given the function: f(x) = x^2 + 1 / x^2 - 9 a)find y and x ...
- Calculus - Given the function: f(x) = x^2 + 1 / x^2 - 9 a)find y and x ...