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What is the number of extremes of the function f(x)=x/(x^2+1)?
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To determine the number of extremes of the function, we'll have to find out how many critical values the function does have.
The critical values of the function are the zeroes of the first derivative of the function.
We'll determine the 1st derivative of the function, using the quotient rule:
f'(x) = [x'*(x^2 + 1) - x*(x^2 + 1)']/(x^2 + 1)^2
f'(x) = (x^2 + 1 - 2x^2)/(x^2 + 1)^2
We'll combine like term inside brackets:
f'(x) = (-x^2 + 1)/(x^2 + 1)^2
We'll cancel f'(x):
(-x^2 + 1)/(x^2 + 1)^2 = 0
Since the denominator is always positive for any x, we'll cancel only the numerator:
-x^2 + 1 = 0
-x^2 = -1
x^2 = 1 => x1 = 1 and x2 = -1
Since the 1st derivative has two zeroes, therefore the function has two critical values.
The number of extremes of the function is of 2.
Posted by giorgiana1976 on July 18, 2011 at 3:48 PM (Answer #1)
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