# `f(x) = 2 + 2x^2 - x^4` (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...

`f(x) = 2 + 2x^2 - x^4` (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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### 2 Answers

Given: `f(x)=2+2x^2-x^4`

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

`f'(x)=4x-4x^3=0`

`4x(1-x^2)=0`

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

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

Part a)

If f'(x)>0 the function is increasing in the interval.

If f'(x)<0 the function is decreasing in the interval.

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

Since f'(-2)>0 the function is increasing in the interval (-`oo` ,-1).

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

Since f'(-1/2)<0 the function is decreasing in the interval (-1, 0).

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

Since f'(1/2)> the function is increasing in the interval (0, 1).

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

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

Part b)

Because the function changes direction from increasing to decreasing a local maximum exist at x=-1 and x=1. The local maximum occurs at the points

(-1, 3) and (1, 3).

Because the function changes direction from decreasing to increasing a local minimum exists at x=0. The local minimum occurs at the point (0, 2).

Part c)

Find the critical values of the second derivative.

`f''(x)=4-12x^2=0`

`4(1-3x^2)=0`

`1=3x^2`

`x=+-(sqrt(3)/3)=+-.577`

If f''(x)>0 the function is concave up.

If f''(x)<0 the function is concave down.

If f"(x)=0 an inflection point exists.

Select an x value in the interval (-`oo` ,-.577).

Since f''(-1)<0 the function is concave down in the interval (-`oo` ,-.577).

Select an x value in the interval (-.577, .577).

Since f''(0)>0 the function is concave up in the interval ('.577, .577)

Select an x value in the interval (.577, `oo` ).

Since f''(1)<0 the function is concave down in the interval (.577, `oo` ).

Since f''( `+-` .577)=0 the function will have inflection points at x=`+-` .577.

The inflection points are at the coordinates (-.577, 2.555) and (.577,2.555).

Part d)

Given

`f(x)=2+2x^2-x^4`

`f'(x)=4x-4x^3` and `f''(x)=4-12x^2`

(a) The function increases when `f'(x)gt0` and decreases when `f'(x)lt0` , so

` 4x-4x^3gt0`

Increases from `(-oo,-1)` and `(0,1)`

`4x-4x^3lt0`

`-1ltxlt0` => decreases from `(-1,0)` and `(1,oo)`

b) Set f'(x)=0

`4x-4x^3=0`

For `xlt-sqrt(1/3) f''(x)lt0` so the function is concave down.

For `xgt-sqrt(1/3) f''(x)gt0` so the function is concave up.

For `xltsqrt(1/3) f''(x)gt0` so the function is concave up.

For `xgtsqrt(1/3) f''(x)lt0` so the function is concave down.

Therefore, `x=+-sqrt(1/3)` are both inflection points.

d) graph