Question

What are the absolute extrema of `f(x)= x^5 -x^3+x^2-7x in [0,7]`?

Answer 1

Minimum: `f(x) = -6.237` at `x= 1.147`

Maximum: `f(x) = 16464` at `x = 7`

Explanation:

We're asked to find the global minimum and maximum values for a function in a given range.

To do so, we need to find the critical points of the solution, which can be done by taking the first derivative and solving for `x`:

`f'(x) = 5x^4 - 3x^2 + 2x - 7`

`x ~~ 1.147`

which happens to be the only critical point.

To find the global extrema, we need to find the value of `f(x)` at `x=0`, `x = 1.147`, and `x=7`, according to the given range:

• `x = 0`: `f(x) = 0`

• `x = 1.147`: `f(x) = -6.237`

• `x = 7`: `f(x) = 16464`

Thus the absolute extrema of this function on the interval `x in [0, 7]` is

Minimum: `f(x) = -6.237` at `x = 1.147`

Maximum: `f(x) = 16464` at `x = 7`