Let (x) = x^3 - 6 x. Find the absolute extrema on (1, 3). a. Find the first derivative and the...

Question:

Let {eq}(x) = x^3 - 6 x {/eq}. Find the absolute extrema on {eq}(1,\ 3) {/eq}.

a. Find the first derivative and the critical points.

(b) Find the maximum and minimum values of {eq}f {/eq} over the interval {eq}[1,\ 3] {/eq}.

A derivative of Function:

To find the critical points of the function we will differentiate it and put it equal to 0 and also those points we will reject which lie outside the interval.

Answer and Explanation:

To find the derivative we will differentiate the equation:

{eq}f(x)=x^{3}-6x {/eq}

Differentiating it we get:

{eq}f'(x)=3x^{2}-6\\ 3x^{2}-6=0\\ x=\pm \sqrt{2} {/eq}

Now let us find the maxima and the minima values at the points:

{eq}f(1)=-5\\ f(3)=9\\ f(\sqrt{2})=-4\sqrt{2} {/eq}

Maxima occurs at x=3

Minima occurs at {eq}x=\sqrt{2} {/eq}


Learn more about this topic:

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Solving Partial Derivative Equations

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 1
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