Question

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

Answer 1

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]}