Find interval where function is decreasing and concave up

  • Thread starter Painguy
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Homework Statement


On what intervals is the function f(x) = x^4 − 4x^3 both decreasing and concave up?
x < 0, 2 < x < 3
x < 3
x < 0, x > 2
x > 0, x < -3
2 < x < 3

Homework Equations




The Attempt at a Solution


I tried to get the derivative of f(x) to make it greater than 0
f'(x)=4x^3 -12x^2
0<4x^3-12x^2
0<3x^2(x-3)
o then take 3x^2 to the other side and end up with
0<x-3
3<x

I tried to get the concavity by taking the double derivative of f(x)
f''(x)=12x^2-24x
0<12x^2-24x
0<12x(x-2)
0<x-2
2<x

so the answer is 2<x<3?
 

Answers and Replies

  • #2
SammyS
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Homework Statement


On what intervals is the function f(x) = x^4 − 4x^3 both decreasing and concave up?
x < 0, 2 < x < 3
x < 3
x < 0, x > 2
x > 0, x < -3
2 < x < 3

Homework Equations




The Attempt at a Solution


I tried to get the derivative of f(x) to make it greater than 0
Why greater than zero. Would that mean the function has positive slope or would it have negative slope? Would this mean that it is increasing or that it is decreasing?
f'(x)=4x^3 -12x^2
0<4x^3-12x^2
0<3x^2(x-3)
o then take 3x^2 to the other side and end up with
0<x-3
3<x

I tried to get the concavity by taking the double derivative of f(x)
f''(x)=12x^2-24x
0<12x^2-24x
0<12x(x-2)
0<x-2
2<x

so the answer is 2<x<3?
 
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