Home > Calculus > Identifying Turning Points (Local Extrema) for a Function

What are the absolute extrema of #f(x)=(x^2 - 1)^3 in[-oo,oo]#?

Answer 1

#6x(x^2-1)#

To compute critical points of a function, we need to compute the first derivative, and then to find its zeroes.

As a general rule, we have that

#d/dx (f(x))^n = n (f(x))^{n-1} * f'(x)#.

Of course, #n=3#, and #f(x)=x^2-1#, which means that #f'(x)=2x#. Plugging all these things into the general formula, we have

#d/dx (x^2-1)^3 = 3(x^2-1)^2 * 2x#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Aurora Alvarado

The absolute extrema of ( f(x) = (x^2 - 1)^3 ) on the interval ([- \infty, \infty]) are as follows:

The function has a local minimum at ( x = -1 ) and a local maximum at ( x = 1 ). Since the function approaches infinity as ( x ) approaches infinity or negative infinity, there are no absolute minima. The absolute maximum occurs at ( x = 1 ), where ( f(x) = 0 ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.