`h(x) = 5x^3 - 3x^5` (a) FInd the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use the...

`h(x) = 5x^3 - 3x^5` (a) FInd the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use the information from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one.

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Textbook Question

Chapter 4, 4.3 - Problem 38 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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mathace | (Level 3) Assistant Educator

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Given: `h(x)=5x^3-3x^5`

Find the critical numbers by setting the first derivative equal to zero and solving for the x values.

`h'(x)=15x^2-15x^4=0`

`15x^2(1-x^2)=0`

`x=0,x=1,x=-1`

The critical values are x=0, x=1, x=-1.

Part a)

If h'(x)>0 the function increases in the interval.

If h'(x)<0 the function decreases in the interval.

Select an x value in the interval (-`oo`  ,-1).

Since h'(-2)<0 the function is decreasing in the interval (-`oo`  ,-1).

Select an x value in the interval (-1, 0).

Since h'(-1/2)>0 the function increases in the interval (-1, 0).

Select an x value in the interval (0, 1).

Since h'(1/2)>0 the function increases in the interval (0, 1).

Select an x value in the interval (1, `oo` ).

Since h'(2)<0 the function decreases in the interval (1,`oo` ).

Because the function changes direction from decreasing to increasing a local minimum exists at x=-1. The local minimum occurs at the point (-1, -2).

Because the function changes direction from increasing to decreasing a local maximum exists at x=1. The local maximum occurs at the point (1, 2).

Part c

`h''(x)=30x-60x^3=0`

`30x(1-2x^2)=0`

`x=0, x=+-(sqrt(2))/(2)=+-.707`

The critical values of the second derivative are x=0, x=-.707, and x=.707

If h''(x)>0 the graph of the function is concave up.

If h''(x)>0 the graph of the function is concave down.

If h''(x)=0 an inflection point exists.

Select an x value in the interval (-`oo` , -.707).

Since h''(-1)>0 the function is concave up in the interval (-`oo`  ,-.707).

Select an x value in the interval (-.707, .707).

Since h''(0)=0 an inflection point will occur at x=0. The inflection point occurs at the coordinate (0, 0).

Select an x value in the interval (.707, `oo` ).

Since h''(1)<0 the function is concave down in the interval (.707, `oo` ).

Part d)

 

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scisser | (Level 3) Honors

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Given `h(x)=5x^3-3x^5`

`h'(x)=15x^2-15x^4`

`h''(x)=30x-60x^3`

a) The function increases when f'(x)>0 and decreases when f'(x)<0, so

`15x^2-15x^4lt0` decreases `(-oo,-1) and (1,oo)`

`15x^2-15x^4gt0` increases `(-1,0) and (0,1)`

b) Set the first derivative equal to 0.

`15x^2-15x^4=0 `

`x=0 or x=1 or x=-1`

Since the function decreases to the left of -1 and increases to the right, x=-1 is a minimum. Also, since the function increases to the left of 1 and decreases to the right, x=1 is a maximum.

c) Set the second derivative equal to 0

`30x-60x^3=0`

`x=0 or x=+-sqrt(1/2)`

For x<0 f''(x)>0 so the function is concave up.

For x>0 f''(x)<0 so the function is concave down.

 

For `xlt-sqrt(1/2) f''(x)gt0` so the function is concave up.

For `xgt-sqrt(1/2) f''(x)lt0` so the function is concave down.

 

For `xltsqrt(1/2) f''(x)gt0` so the function is concave up.

For `xgtsqrt(1/2) f''(x)lt0` so the function is concave down.

Therefore, all the points are inflection points

d) graph

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