BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4, Problem 15RE
To determine

Tofind:the interval of increase and decrease function, interval of concavity, local maxima,local minima and inflection points.

Expert Solution

Answer to Problem 15RE

The function increases in the interval of (3,0),(0,3)

The function decreases in the interval of (,3),(3,) .

Local maximum: f(3)=293

Local minima: f(3)=293 .

Concave upward at (6,0),(6,) .

Concave downward at (,6),(0,6) .

IP lies in the interval of (6,5366),(6,5366) .

Explanation of Solution

Given:

  f(x)=x2-1x3 .

Concept used:

Increasing or decreasing function can be calculated by equating first derivative of the function to 0.

  dydx=0 .

If d2ydx2>0. the concave will be open upward and local minima can be found.

If d2ydx2>0. the concave will open downward and local maxima can be found.

If the graph of y=f(x) has point of inflection then y=f(x)=0 .

Calculation:

  f(x)=x2-1x3f(x)=x3.2x(x21).3x2(x3)2.2x4(3x43x2)x6=x4+3x2x6.f(x)=x2+3x4=0.x=±3.

The function increases in the interval of

  (3,0),(0,3) .

The function decreases in the interval of

  (,3),(3,) .

Local maximum:

  f(3)=293 .

Local minima:

  f(3)=293.2x212=0.x=±6.f(x)=x4(2x)(x2+3)4x3(x4)2=2x5(4x5+12x3)x8=2x512x3x8=2x212x5.

Concave upward at (6,0),(6,) .

Concave downward at (,6),(0,6) .

Inflection point lies in the interval of:

  f(x)=x2-1x3

where x=±6 .

Put the value of x here to get the value of y coordinate:

  f(6)=5366.f(6)=5366.

IP lies in the interval of (6,5366),(6,5366) .

Hence,

The function increases in the interval of (3,0),(0,3).

The function decreases in the interval of (,3),(3,) .

Local maximum: f(3)=293

Local minima: f(3)=293 .

Concave upward at (6,0),(6,) .

Concave downward at (,6),(0,6) .

IP lies in the interval of (6,5366),(6,5366) .

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