   Chapter 3.5, Problem 7E

Chapter
Section
Textbook Problem

# 1-40 Use the guidelines of this section to sketch the curve. y = 1 5 x 5 − 8 3 x 3 + 16 x

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=15x5-83x3+16x

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 y=15x5-83x3+16x  is a polynomial, its domain is -,  .

B) Intercepts

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

y=1505-8303+160

y=0

y intercept is 0, 0.

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

0=15x5-83x3+16x

x=0

x  intercept is 0, 0.

C) Symmetry

To find the axis of symmetry, replace x by -x.

Therefore,

f-x=15(-x)5-83(-x)3+16(-x)

f-x= -15x5-83x3+16x= -fx

It is an odd function, so it has rotational symmetry about the origin.

Since it’s a polynomial, it’s not periodic.

D) Asymptote

This is a polynomial function. Therefore, it has no asymptote.

E) Intervals of increase or decrease

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

f'x= x4-8x2+16

Equating this derivative with 0; therefore, x4-8x2+16=0 .

Solve this for x, and get critical points as x=-2 and x=2.

By using these critical points and the domain, create three intervals as - , -2, -2, 2  & (2 ,  )

Now, take a test point from each of the above intervals, and check whether the function is increasing or decreasing in that interval.

For - ,-2, consider x= -3.

f'(-3)=-34-8-32+16

f'-3=25

-<x<-2;, f'x>0

The function is increasing in the interval (-,-2).

For -2, 2, consider x= 0.

f'0=04-8(0)2+16  f'0=16

-2<x<2,  f'x>0.

The function is increasing in the interval -2, 2

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