Chapter 4: Q. 4 (page 193)

Use graphing utility to approximate (round to two decimal places) the local maximum value and local minimum value of $f (x) = x^{3} - 2 x^{2} - 4 x + 5$ , for $- 3 < x < 3$ .

Short Answer

Expert verified

The solution is the local maximum value is $6.48$ at $x = - 0.67$ and local minimum value is $- 3$ at $x = 2$

Step by step solution

Step 1. Given information

Function: $f (x) = x^{3} - 2 x^{2} - 4 x + 5$ for $- 3 < x < 3$

Step 2. Calculation

Graph of the function $f (x) = x^{3} - 2 x^{2} - 4 x + 5$ for $- 3 < x < 3$ is -

So, the local maximum value is $6.48$ at $x = - 0.67$ and local minimum value is $- 3$ at $x = 2$ .

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Use graphing utility to approximate (round to two decimal places) the local maximum value and local minimum value of $f (x) = x^{3} - 2 x^{2} - 4 x + 5$ , for $- 3 < x < 3$ .

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation

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Use graphing utility to approximate (round to two decimal places) the local maximum value and local minimum value of fx=x3-2x2-4x+5, for -3<x<3.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

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Study anywhere. Anytime. Across all devices.

Use graphing utility to approximate (round to two decimal places) the local maximum value and local minimum value of $f (x) = x^{3} - 2 x^{2} - 4 x + 5$ , for $- 3 < x < 3$ .