What are the local extrema, if any, of #f (x) = x^4-8x^2-48 #?

1 Answer
Dec 15, 2016

Answer:

Local maximum at #(0,-48)#
Local minimum at #(-2,-56)# and #(2,-56)#

Explanation:

# f(x)=x^4-8x^2-48 #

Local extrema occur when #f'(x)=0#

Differentiating wrt #x# we get:

# f'(x)=4x^3-16x #

So if #f'(x)=0#
# => 4x^3-16x = 0#
# :. \ 4x(x^2-4) = 0#
# :. \ 4x(x^2-2^2) = 0#
# :. \ 4x(x+2)(x-2) = 0#
# :. \ x=0, +-2#

We can now examine the nature of these turning points by looking at the second derivative:

Differentiating #f'(x)# wrt #x# we get:

# f''(x)=12x^2-16 #

so # { (x=+-2,=> f''(x)>0, "ie minimum"), (x=0,=> f''(x)<0, "ie maximum") :} #

And the values of the function at these extrema are:

#f(0) \ \ \ \ \ = 0 - 0 - 48 \ \ \ = -48#
#f(+-2) = 16 - 24 - 48 = -56#

Hence the extrema are:

Local maximum at #(0,-48)#
Local minimum at #(-2,-56)# and #(2,-56)#

We can confirm this visually by plotting the graph:
graph{x^4-8x^2-48 [-10, 10, -74.2, 74]}