   Chapter 10.2, Problem 15E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Find the relative maxima, relative minima, and points of inflection and sketch the graphs of the functions in Problems 15-20. y = 1 3 x 3 − 2 x 2 + 3 x + 2

To determine

To calculate: The relative minimum, relative maximum and points of infection for the provided function y=13x32x2+3x+2.

Explanation

Given Information:

The provided function is,

y=13x32x2+3x+2

Formula used:

To find relative maxima and minima of a function,

1. Find the critical values of a function.

2. Substitute the critical values into f(x) to find the critical points.

3. Evaluate f(x) at each critical value for which f(x)=0.

(a) If f(x0)<0, a relative maximum occurs at x0.

(b) If f(x0)>0, a relative minimum occurs at x0.

(c) If f(x0)=0 or f(x0) is undefined, the second derivative test fails and then use the first derivative test.

Calculation:

Consider the provided function,

y=13x32x2+3x+2

Now, obtain the first derivative as,

y=13x32x2+3x+2y=x24x+3=(x3)(x1)

Now, to obtain the critical values, set y=0 as,

(x3)(x1)=0x=3

And,

x=1

Thus, the critical points are at x=3 and x=1.

Now, obtain the second derivative as,

y=13x32x2+3x+2

Differentiate the function with respect to x,

y=x24x+3

Differentiate the function with respect to x,

y=2x4

To obtain point of inflection, set y=0 as,

2x4=02x=4x=2

Thus, x=2 is the point of inflection.

Now, substitute 3 for x in the y as,

y=2(3)4=64=2

Since, y(3)>0, so relative minimum occurs at x=3.

Now, substitute the critical point 1 for x in y as,

y=2(1)4=24=2

Since, y(1)<0, so relative maximum occurs at x=1

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