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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Produce graphs of f that reveal all the important aspects of the curve. In particular, you should use graphs of f′ and f″ to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points.

f(x) = x6 − 5x5 + 25x3 − 6x2 − 48x

To determine

To sketch: The graph of f that reveals all important aspects of the curve and the graph of f' and f" to estimate the intervals of increase, decrease, the extreme values, the intervals of concavity and the inflection points.

Explanation

Given information:

The curve function is f(x)=x65x5+25x36x248x (1)

Calculation:

Equate Equation (1) to 0.

f(x)=0x65x5+25x36x248x=0 (2)

Solve Equation (2).

x=0,3.20

Draw the graph of the function f by substituting different values for x.

The sketch of the function f is shown in Figure 1.

Differentiate Equation (1) with respect to x.

f'(x)=6x525x4+75x212x48 (3)

Equate Equation (3) to 0.

f'(x)=06x525x4+75x212x48=0 (4)

Solve Equation (4).

x=1.31,0.84,1.06,2.50,2.75

The sketch of the function f' is shown in Figure 2.

Differentiate Equation (3) with respect to x.

f"(x)=30x4100x3+150x12 (5)

Equate Equation (5) to 0.

f"(x)=030x4100x3+150x120=0 (6)

Solve Equation (6).

x=1.1,0.08,1.71,2.63

The sketch of the function f" is shown in Figure 3.

Refer Figure 2.

The curve f is decreasing on (,1.31), (0.84,1.06), and (2.50,2.75).

The curve f is increasing on (1.31,0.84), (1.06,2.50), and (2.75,).

The local minimum value occurs at x=1.31, x=1.06, and x=2.75.

The local maximum value occurs at x=0.84 and x=2.50.

Therefore, the intervals of increase are (1.31,0.84)_, (1.06,2.50)_, and (2.75,)_.

Therefore, the intervals of decrease are (,1.31)_ (0.84,1.06)_, and (2.50,2.75)_.

Find the extreme values (local maxima and local minima) as shown below:

Substitute 1.31 for x in Equation (1).

f(1.31)=(1.31)65(1.31)5+25(1.31)36(1.31)248(1.31)=20.72

Substitute 1.06 for x in Equation (1).

f(1.06)=(1.06)65(1.06)5+25(1.06)36(1.06)248(1.06)=33.12

Substitute 2.75 for x in Equation (1).

f(2.75)=(2.75)65(2.75)5+25(2.75)36(2

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