F's question at Yahoo Answers involving absolute extrema

[MarkFL]

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]

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:

[math]f(x)=x^3−2x^2−4x+8[/math]

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

[math]f'(x)=3x^2-4x-4=(3x+2)(x-2)=0[/math]

Hence, the critical values are:

[math]x=-\frac{2}{3},\,2[/math]

The critical points are then:

[math]\left(-\frac{2}{3},f\left(-\frac{2}{3} \right) \right)=\left(-\frac{2}{3},\frac{256}{27} \right)[/math]

[math](2,f(2))=(2,0)[/math]

The end-point values are:

[math](-1,f(-1))=(-1,9)[/math]

[math](3,f(3))=(3,5)[/math]

Since [math]0<5<9<\frac{256}{27}[/math] we may state that:

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

Absolute maximum occurs at [math]\left(-\frac{2}{3},\frac{256}{27} \right)[/math]

To F and any other guests viewing this topic, I invite and encourage you to post other absolute extrema questions here in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.