   Chapter 4.5, Problem 6E

Chapter
Section
Textbook Problem

Use the guidelines of this section to sketch the curve.y = x5 − 5x

To determine

To Find: The curve by plotting x and y coordinates using guideline.

Explanation

Given:

The given curve is

f(x)=y=(x55x) (1)

f(x)=y=x(x45) (2)

Calculation:

(a)

Calculate the domain.

From the equation (1), the function is defined for all real values of x (there is no restriction on the value of x). Hence, the domain of f(x) is all real values of x .

For the polynomial the domain is (,) .

(b)

Calculate the intercepts.

Calculate the y –intercept.

f(x)=y=(x55x)

f(0)=(050x)=0

Calculate the value of x-intercept.

Substitute 0 for y in the equation (2).

x(x45)=yx(x45)=0

x=0 or (x45)=0

x=0 or x=±514

Therefore, the x -intercept are x=(0,±514) .

(c)

Calculate the symmetry.

Substitute x for x in the equation (1)

f(x)=(x55x)f(x)=(x55(x))f(x)=(x5+5x)

f(x)=(x55x)

f(x)=f(x)

It is an odd function.

So, it has rotational symmetry about the origin

(d)

Calculate asymptotes.

Apply limit of x tends to (x) in the equation (1).

limx(x55x)=(55)=

Apply limit of x tends to (x) in the equation (1).

limx(x55x)=(55())=

Here, the value of limit gets infinity, so there is no horizontal and vertical asymptote.

(e)

Calculate the intervals.

Differentiate the equation (1) with respect to x .

f(x)=(x55x)

f'(x)=5x45 (2)

Consider the limit (,1) .

Substitute the -2 for the value of x in (2)

f'(x)=5(2)45=805f'(x)=75>0

Consider the limit (1,1)

Substitute the 0 for the value of x in (2)

f'(x)=5(0)45=5<0

Consider the limit (1,)

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