points of inflection?

Question

Find all points of inflection for f(x)=x^4-6x^3...

also give a sign of the second derivative of f at the indicated poin/points in the problem (positive, negative, zero)

Answer 1

f(x)=x^4-6x^3
f'(x)= 4x^3 -18x^2
f''(x)= 12x^2 - 36x

To find points of inflection, we use the second derivative (f''[x]=0)
So f''(x)= 12x^2 - 36x=0
x(12x-36)=0
x=0, 36/12 (3)

_+_|_--_|_+_
......0.....3.......
Therefore, 0 and 3 are inflection points
Concave up between (-infinity,0), (3, infinity)
Concave down between (0,3)

Answer 2

Just a reminder on the concept: points of inflection occur when the second derivative changes sign, or in other words, when your function changes from concave up to concave down. You find these points by determining where the second derivative equals zero and then checking to see if there is a sign change at any of these points.

Solution:

Find second derivative of f(x): 12x^2 - 36x

Set equal to zero and solve: 12x^2 - 36x = 0
x = 0 and x = 3

Set up intervals of concavity and check signs:

(negative infinity, 0) (0,3) (3, infinity)
f''(x)>0 f(x)<0 f(x)>0
Concave up concave down concave up

[Thus, the second derivative changes sign at x=0 and x=3]

Find y-coordinates (if necessary):

f(0) = 0
f(3) = -81

So the points are (0,0) and (3, -81)

Good luck!

Answer 3

Inflection points occur when the second derivative changes from + to - or vice-versa. So we find when they are 0, then check a bit before and after to make sure they do change there.

f'(x)=4x^3-18x^2
f''(x)=12x^2-36x

0=12x^2-36x
0=x(12x-36)
x=0 or x=3
f''(0)=0, f''(3)=0

Let's try a bit before and after 0, and then 3, just to be sure.

f''(0.1)=12(0.1)^2-36(0.1)<0
f''(-0.1)=12(-0.1)^2-36(-0.1)>0

So 0 is an inflection point (f''(x) changes from + to - at x=0). Now we check 3:

f''(3.1)=12(3.1)^2-36(3.1)>0
f''(2.9)=12(2.9)^2-36(2.9)<0

So 3 is also an inflection point (f''(x) changes from - to + at x=3).

So the two points are (0,f(0)),(3,f(3)), or (0,0) and (3,-81). The second derivative is ALWAYS 0 at inflection points.

There's your answer.

Answer 4

points of inflection occur when f''=0 and there is a change in sign of f'' if you evaluate f'' just to the left and just to the right of the potential point of inflection

for the function above:

f'=4x^3-18x^2
f''=12x^2-36x

f''(x)=0 when x=0 and when x=3

the value of f'' for x just less than 0 is positive, and for x slightly greater than 0 is negative, so x=0 is a point of inflection

the value of f'' for x just greater than 3 is positive, and is negative for x just less than 3, so x=3 is also a point of inflection

Answer 5

f'=4x^3-18x^2
f''=12x^2-36x


f''>0 when 12x^2>36x