Thread: Concavity

a)Determine the concavity and points of inflection.
b) Find the local max and local min points.

Can someone check my work?

f(x)=8-x^(1/3)
f'(x)=-1/(3^(2/3))
f'(x)= undefined when x=0
(0,8)

f''(x)=2/(9x^(5/3))
f''(x)=undefined when x=0
(0,8)

therefore point of inflection occurs at (0,8) and there is no concavity, no local max and min points.

Question: for this particular case, a vertical tangent, there is no concavity correct??

Originally Posted by skeske1234

therefore point of inflection occurs at (0,8) and there is no concavity, no local max and min points.

Question: for this particular case, a vertical tangent, there is no concavity correct??

I agree with this

Originally Posted by skeske1234

f(x)=8-x^(1/3)
f'(x)=-1/(3^(2/3))
f'(x)= undefined when x=0
(0,8)

f''(x)=2/(9x^(5/3))
f''(x)=undefined when x=0
(0,8)

I don't really know what you have done here.

Originally Posted by skeske1234

a)Determine the concavity and points of inflection.
b) Find the local max and local min points.

Can someone check my work?

f(x)=8-x^(1/3)
f'(x)=-1/(3^(2/3))
f'(x)= undefined when x=0
(0,8)

f''(x)=2/(9x^(5/3))
f''(x)=undefined when x=0
(0,8)

therefore point of inflection occurs at (0,8) and there is no concavity, no local max and min points.

Question: for this particular case, a vertical tangent, there is no concavity correct??

f''(x) < 0 for x < 0 ... f(x) is concave down on this interval

f''(x) > 0 for x > 0 ... f(x) is concave up on this interval

f(x) changes concavity at x = 0, therefore it is a point of inflection.

ok, so let me re-try this:

a) point of inflection (0,8)
concave up when x>0
concave down when x<0

b) no local max or min points

is this correct now?

and another question on a side note: when a function has a vertical tangent, that means that the function can still have concavity, right?

Can someone please answer and verify the above? thanks