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

Jul 2, 2017

Minimum: $f \left(x\right) = - 6.237$ at $x = 1.147$

Maximum: $f \left(x\right) = 16464$ 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 $x$:

$f ' \left(x\right) = 5 {x}^{4} - 3 {x}^{2} + 2 x - 7$

$x \approx 1.147$

which happens to be the only critical point.

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

• $x = 0$: $f \left(x\right) = 0$

• $x = 1.147$: $f \left(x\right) = - 6.237$

• $x = 7$: $f \left(x\right) = 16464$

Thus the absolute extrema of this function on the interval $x \in \left[0 , 7\right]$ is

Minimum: $f \left(x\right) = - 6.237$ at $x = 1.147$

Maximum: $f \left(x\right) = 16464$ at $x = 7$