Answered: Given f(x) = 2x° – 39x? + 240x + 3,…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Given f(x) = 2x° – 39x? + 240x + 3, answer the following:
a) The point where the relative minimum occurs:
b) The point where the relative maximum occurs:

Expert Solution

Step 1

Given: $f (x) = 2 x^{3} - 39 x^{2} + 240 x + 3$

To determine:

a) The point where the relative minimum occurs.

b) The point where the relative maximum occurs.

Step 2

Calculation:

$f (x) = 2 x^{3} - 39 x^{2} + 240 x + 3$

Differentiate $f (x)$ w.r.t $x$

$f' (x) = 6 x^{2} - 78 x + 240$

Put $f' (x) = 0$

$\Rightarrow 6 x^{2} - 78 x + 240 = 0 \Rightarrow x^{2} - 13 x + 40 = 0 \Rightarrow x^{2} - 8 x - 5 x + 40 = 0 \Rightarrow x (x - 8) - 5 (x - 8) = 0 \Rightarrow (x - 5) (x - 8) = 0 \Rightarrow x = 5, x = 8$

These are the critical points.

Now differentiate $f' (x)$ w.r.t $x$

$f'' (x) = 12 x - 78$

At $x = 5$

$f'' (5) = 12 (5) - 78 = - 18 < 0$

$∴ x = 5$ is the point of relative maxima.

At $x = 8$

$f'' (8) = 12 (8) - 78 = 18 > 0$

$∴ x = 8$ is the point of relative minima.

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Given f(x) = 2x° – 39x? + 240x + 3, answer the following: a) The point where the relative minimum occurs: b) The point where the relative maximum occurs: