Calculus

posted by .

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 -

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.

Similar Questions

1. math

Determine whether this function has a mzximum or minimum value and then find that value: f(x)=2x^2 + 3x -9 The derivative, f'(x) = 4x +3, is zero when x = -3/4 . That means there is a relative extreme value (either maximum or minimum) …
2. Calculus - another question

"Find the absolute minimum value of the function f(x) = x ln(x)". I know that one of the critical values is 1 (domain restriction). Am I right?
3. Calculus

For y=(1/4)x^4-(2/3)x^3+(1/2)x^2-3, find the exact intervals on which the function is a. increasing b. decreasing c. concave up d. concave down Then find any e. local extreme values f. inflection points --- I thought it'd be better …

1. Find all critical values for f(x)=(9-x^2)^⅗ A. 0 B. 3 C. -3,3 D. -3, 0, 3 E. none of these I got D. I found the derivative and solved for critical numbers. 2. Find all intervals on which the graph of f(x)=(x-1)/(x+3) is concave …

1. Find all points of inflection: f(x)=1/12x^4-2x^2+15 A. (2, 0) B. (2, 0), (-2, 0) C. (0, 15) D. (2, 25/3), (-2, 25/3) E. none of these I got D. I found the second derivative and equaled it to 0 and solved for x. I plugged the x values …

1. Find all points of inflection of the function f(x)=x^4-6x^3 A. (0, 0) B. (0, 0), (9/2, -2187/16) C. (3, -81) D. (0, 0), 3, -81) E. none of these I got D. I found the second derivative and solved for x and plugged values into original …