# F's question at Yahoo! Answers involving absolute extrema

#### MarkFL

Staff member
Here is the question:

Math Question - Calculus!!?

Find the absolute maxima and minima for f(x) on the interval [a, b].
f(x) = x3 − 2x2 − 4x + 8, [−1, 3]

absolute maximum
(x, y) =

absolute minimum
(x, y) =
Here is a link to the question:

Math Question - Calculus!!? - Yahoo! Answers

I have posted a link there so the OP can find my response.

#### MarkFL

Staff member
Hello F,

The absolute extrema of a function on some interval can occur either at the critical values, or at the end-points of the interval.

To determine the critical values of the given function:

$$\displaystyle f(x)=x^3−2x^2−4x+8$$

we equate the first derivative to zero, and solve for $x$:

$$\displaystyle f'(x)=3x^2-4x-4=(3x+2)(x-2)=0$$

Hence, the critical values are:

$$\displaystyle x=-\frac{2}{3},\,2$$

The critical points are then:

$$\displaystyle \left(-\frac{2}{3},f\left(-\frac{2}{3} \right) \right)=\left(-\frac{2}{3},\frac{256}{27} \right)$$

$$\displaystyle (2,f(2))=(2,0)$$

The end-point values are:

$$\displaystyle (-1,f(-1))=(-1,9)$$

$$\displaystyle (3,f(3))=(3,5)$$

Since $$\displaystyle 0<5<9<\frac{256}{27}$$ we may state that:

Absolute minimum occurs at $(2,0)$.

Absolute maximum occurs at $$\displaystyle \left(-\frac{2}{3},\frac{256}{27} \right)$$

To F and any other guests viewing this topic, I invite and encourage you to post other absolute extrema questions here in our Calculus forum.

Best Regards,

Mark.