Question

Find the intervals on which the following function is increasing and decreasing and as well as its concavity? f(x)=x^2(1+x^2)

I can derive the f' and f" but I'm having difficulties finding the values
f'(x)=2x+4x^3
f"(x)=2+12x^2

Do I just evaluate the f' and f" without equation the functions to 0?

Answer 1

`f(x)` is decreasing for `x<0`
`f(x)` is increasing for `x>0`
`f(x)` is concave upward for all `x`.

Explanation:

The question is asking you to find the intervals for which `f'(x)` and `f''(x)` are positive and negative.

As you probably already know:

If `f'(x)` > 0, `f(x)` is increasing at x.
If `f'(x)` = 0, `f(x)` has a horizontal tangent at x.
If `f'(x)` < 0, `f(x)` is decreasing at x.

If `f''(x)` > 0, `f(x)` is concave upward at x.
If `f''(x)` = 0, `f(x)` has an inflection point at x.
If `f''(x)` < 0, `f(x)` is concave downward at x.

`f(x)=x^2(1+x^2)=x^4+x^2`
`f'(x)=4x^3+2x`
`f''(x)=12x^2+2`

First let's evaluate `f'(x)`

`f'(x)=4x^3+2x=2x(2x^2+1)`

By examining the factored form of `f'(x)` we can see that `(2x^2+1)` is positive for all real values of x.

Therefore, `f'(x)` is only negative if `2x<0->x<0`.

`f(x)` is decreasing for `x<0`
`f(x)` is increasing for `x>0`

Now let's examine `f''(x)`.

`f''(x)=12x^2+2`

Once again by examination, we can see that `f''(x)` is positive for all real values of x. Therefore:

`f(x)` is concave upward for all `x`.

As visual proof, let's look at a graph of `f(x)`:

graph{x^4+x^2 [-2.574, 2.425, -0.33, 2.17]}

As expected, it is a parabola centred at `x=0`.