What are the local extrema, if any, of #f (x) =80+108x-x^3 #?

Question

Aurora Alvarado · Accepted Answer

We have a local minima at #x=-6# and a local maxima at #x=6#.

A maxima is a high point to which a function #f(x)# rises and then falls again. As such the slope of the tangent or the value of derivative #f'(x)# at that point will be zero. Further, as the tangents to the left of maxima will be sloping upwards, then flattening and then sloping downwards, slope of the tangent will be continuously decreasing, i.e. the value of second derivative #f''(x)# would be negative. A minima on the other hand is a low point to which a function falls and then rises again. As such the tangent or the value of #f'(x)# at minima too will be zero. But, as the tangents to the left of minima will be sloping downwards, then flattening and then sloping upwards, slope of the tangent will be continuously increasing or the value of second derivative #f''(x)# would be positive As #f(x)=80+108x-x^3# #f'(x)=108-3x^2# and this is zero for #x# given by #3x^2-108=0# or #x^2-36=0# or #x=+-6# Now #f''(x)=-6x# and hence at #x=-6#, #f''(x)=36# so we have a local minima with value #-352# and at #x=6#, #f''(x)=-36# so we have a local maxima with value #512# However, as elsewhere, #f'(x)# takes values less than minima and more than maxima elsewhere, these are local maxima and minima. graph{80+108x-x^3 [-20, 20, -700, 700]}

Axel Alvarez · Answer

undefined To find the local extrema of the function $ f(x) = 80 + 108x - x^3 $, first, find the derivative of the function, then set it equal to zero and solve for $ x $. The critical points obtained will be checked for local extrema using the second derivative test.

First derivative:
$ f'(x) = 108 - 3x^2 $

Setting $ f'(x) = 0 $ to find critical points:
$ 108 - 3x^2 = 0 $
$ 3x^2 = 108 $
$ x^2 = 36 $
$ x = \pm 6 $

Second derivative:
$ f''(x) = -6x $

Test the critical points:
For $ x = 6 $, $ f''(6) = -6(6) = -36 $ (negative), so $ x = 6 $ is a local maximum.
For $ x = -6 $, $ f''(-6) = -6(-6) = 36 $ (positive), so $ x = -6 $ is a local minimum.

Therefore, the local maximum occurs at $ x = 6 $, and the local minimum occurs at $ x = -6 $.