critical points and inflection points

[lauren52]

hello,
I need help in graphing, plotting critical points and inflection points of the function f(x) =x^4+8x^3+12
Thus far, I came up with the following; however, I am unsure if this is correct.
f'(x)=4x^3+24^2
4x^2(x+6)=0
4x^2=no solution
x+6=0 x=-6
critical point is -6

Plugging -6 into the original equation
-6^4+8(-6)^3+12
1296-1728+12=-420

no relative minimum
relative maximum (-6,-420)

using the 2nd derivative to find inflection points
12x^2+48x
12x(x+4)=0
12x=no solution
x=-4
plugging -4 into the original equation
-4^4+8(-4)^3+12=204
Inflection points (-4,204)

 

[skeeter]

hello,
I need help in graphing, plotting critical points and inflection points of the function f(x) =x^4+8x^3+12
Thus far, I came up with the following; however, I am unsure if this is correct.
f'(x)=4x^3+24^2
4x^2(x+6)=0
4x^2=no solution oh no? how about x = 0 ?
x+6=0 x=-6
critical point is -6 and 0

Plugging -6 into the original equation
-6^4+8(-6)^3+12
1296-1728+12=-420

no relative minimum
relative maximum (-6,-420)

using the 2nd derivative to find inflection points
12x^2+48x
12x(x+4)=0
12x=no solution 12x = 0 ... x = 0
x=-4
plugging -4 into the original equation
-4^4+8(-4)^3+12=204
Inflection points (-4,204)

try it again

 

[lauren52]

f'(x)=4x^3+24^2
4x^2(x+6)=0
4x^2=0
x+6=0 x=-6
critical point is -6 and 0

Plugging -6 into the original equation
-6^4+8(-6)^3+12
1296-1728+12=-420
and
Plugging in 0 into the original equation
0^4+8(0)^3+12=12

no relative minimum
relative maximum (-6,-420)

using the 2nd derivative to find inflection points
f''(x)=12x^2+48x
12x(x+4)=0
12x=0, x=-4

plugging 0 into the orginal equation
f(0)=0^4+8(0)^3+12=12
Inflection points (0,12)

f(-4)=4^4+8(4)^3+12=12
Inflection points (-4,12)