Question

What are the global and local extrema of `f(x)=x^3+48/x` ?

Answer 1

Local: `x = -2, 0, 2`
Global: `(-2, -32), (2, 32)`

Explanation:

To find extrema, you just find points where `f'(x) = 0` or is undefined . So:

`d/dx(x^3 + 48/x) = 0`

To make this a power rule problem, we'll rewrite `48/x` as `48x^-1`. Now:

`d/dx(x^3 + 48x^-1) = 0`

Now, we just take this derivative. We end up with:

`3x^2 - 48x^-2 = 0`

Going from negative exponents to fractions again:

`3x^2 - 48/x^2 = 0`

We can already see where one of our extrema will occur: `f'(x)` is undefined at `x = 0`, because of the `48/x^2`. Hence, that is one of our extrema.

Next, we solve for the other(s). To start, we multiply both sides by `x^2`, just to rid ourselves of the fraction:

`3x^4 - 48 = 0`
`=> x^4 - 16 = 0`
`=> x^4 = 16`
`=> x = ±2`

We have 3 places where extrema occur: `x = 0, 2, -2`. To figure out what our global (or absolute) extrema are, we plug these into the original function:

So, our absolute minimum is the point `(-2, -32)`, while our absolute maximum is `(2, -32)`.

Hope that helps :)