   Chapter 3.5, Problem 29E

Chapter
Section
Textbook Problem

# 1-40 Use the guidelines of this section to sketch the curve. y = x − 3 x 1 / 3

To determine

To sketch:

The curve of the given function.

Explanation

1) Concept:

i) A domain is the set of x values that satisfy the function.

ii) To find x-intercept, put y=0, and to find y-intercept, put x=0 in the given function.

iii) Symmetry: To find symmetry, replace x by –x and check the behaviour of function. Thus, if f-x=fx, then it is an even function, so it has y-axis symmetry. If f-x=-fx, then it is an odd function, so it has x-axis symmetry. And if f-x-fxfx, then it has no symmetry.

iv) An asymptote is a tangent at infinity. To find horizontal, vertical, and slant asymptote, follow the rules.

v) A function is increasing if f'x>0  and decreasing if f'x<0 in that particular interval.

vi) The number f(c) is a local maximum value of f  if fcf(x) when x is near c and is a local minimum value of f if fc f(x) when x is near c.

vii) If f''x>0, the function is concave up and if f''x<0, the function is concave down in that particular interval. And if f''x=0, give the values of inflection points.

2) Given:

y=x-3x1/3

3) Calculation:

Here, first find the domain of the given function and the x & y intercepts. Next, check the symmetry, asymptotes, intervals of increase and decrease, local maximum and minimum values, concavity, and points of inflection. Using these, sketch the curve.

A) Domain:

Since the function never becomes undefined, the domain is (-,).

B) Intercepts:

For y intercept, plug x=0  in the given function, and solve it for y.

y= x-3x13

y=0

y  intercept is 0, 0.

For x intercept, plug y=0 in the original function, and solve it for x.

0= x-3x13=x3-27x

0=x(x2-27)

x=0, 33.

x  intercepts are 0, 0, (33, 0).

C) Symmetry:

For symmetry, replace x by -x.f-x= -x+3x13=-(x-3x1/3)

f-x= -f(x)

The function is odd. Symmetry is about the origin.

D) Asymptote

Horizontal asymptotes:

limx-  x-3x1/3=-, limx x-3x1/3=

Horizontal asymptote:

No

Vertical asymptotes:

No vertical asymptote

Slant asymptotes:

No slant asymptotes.

E) Intervals of increase or decrease.

To find the intervals of increase or decrease, find the derivative of the given function.

f'x= 1-x-2/3

Equating this derivative with 0 gives

1-x-2/3 =0

x=±1

Also f'0=0

These are the critical points in the domain.

By using domain and these critical points, create intervals.

So intervals are -,-1, -1, 0, 0, 1,1,

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