# `g(x) = 200 + 8x^3 + 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...

`g(x) = 200 + 8x^3 + 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

`g(x)=200+8x^3+x^4`

differentiating,

`g'(x)=24x^2+4x^3`

`g'(x)=4x^2(x+6)`

Now let us find our critical numbers by setting g'(x)=0

`4x^2(x+6)=0`

solving above , x=0 and x=-6

Now let us check sign of g'(x) at test values in the intervals (-`oo` ,-6) , (-6,0) and (0,`oo` )

`f'(-7)=4(-7)^2(-7+6)=-196`

`f'(-1)=4(-1)^2(-1+6)=20`

`f'(1)=4(1)^2(1+6)=28`

Since f'(-7) is negative , function is decreasing in the interval (-`oo` ,-6)` `

f'(-1) is positive so function is increasing in the interval (-6,0)

f'(1) is positive so function is increasing in the interval (0,`oo` )

Since f'(-7) is negative and f'(-1) is positive , therefore Local minimum is at x=-6 . Local minima can be found by plugging in x=-6 in the function.

`f(-6)=200+8(-6)^3+x^4=-232`

**Local minimum=-232 at x=-6**

Now to find the intervals of concavity and inflection points,

`g''(x)=4(x^2+(x+6)(2x))`

`g''(x)=4(x^2+2x^2+12x)`

`g''(x)=4(3x^2+12x)`

`g''(x)=12x(x+4)`

set g''(x)=0

x=0 , x=-4

Now let us check the concavity in the intervals (-`oo` ,-4) , (-4,0) and (0,`oo` ) by plugging test values in g''(x).

`g''(-5)=12(-5)(-5+4)=60`

`g''(-2)=12(-2)(-2+4)=-48`

`g''(1)=12(1)(1+4)=60`

Since g''(-5) is positive , function is concave up in the interval (-`oo` ` ` ,-4)

g''(-2) is negative , so the function is concave down in the interval (-4,0)

g''(1) is positive , so the function is concave up in the interval (0,`oo` )

Since the concavity is changing so x=0 and x=-4 are the inflection points.

Given `g(x)=200+8x^3+x^4`

`g'(x)=24x^2+4x^3`

`g''(x)=48x+12x^2`

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

`24x^2+4x^3gt0` increases from `(-6,0) and (0,oo)`

`24x^2+4x^3lt0` decreases from `(-oo,-6)`

b) Set the first derivative equal to 0

`24x^2+4x^3=0`

`x=0 and x=-6`

Since the function is decreasing to the left of -6 and increasing to the right, x=-6 is a minimum.

c) set the second derivative equal to 0.

`48x+12x^2=0`

`x=0 and x=-4`

For `xlt0 f''(x)lt0` so the function is concave down.

For `xgt0 f''(x)gt0` so the function is concave up.

For `xlt-4 f''(x)gt0` so the function is concave up.

For `xgt-4 f''(x)lt0` so the function is concave down.

therefore both `x=0 and x=4` are inflection points.

d) graph