f(x)=X^3-6X^2+12X-5?

Question

Find all inflection points & state the intervals where the function is concave up or down

Answer 1

First, find the second derivative:
f' = 3x^2 - 12x + 12
f'' = 6x^2 - 12

Then set the second derivative = 0 and solve

6x^2 - 12 = 0
6x^2 = 12
x^2 = 2
x = sqrt 2 and -sqrt 2

Check points on all sides of that... Let's use x = -2, 0 and 2. Put those into the second derivative:
f''(-2) = 6(-2)^2 - 12 = 12 = positive, so the graph is concave up prior to (-sqrt 2).
f''(0) = -12 = negative, so concave down between (-sqrt 2) and (sqrt 2).
f''(2) = 12 = positive, so concave up after (sqrt 2).

Answer 2

You can find concave intervals using the second derivative of a function.

f'(x)=3x^2-12x+12
f''(x)=6x-12

Set the second derivative equal to 0 and solve:
0=6x-12
6x=12
x=2

Now set up a chart: Test the value of f''(x) in each interval
Interval f''(x) Concavity
(-infty,2) - Concave down
(2, infty) + Concave up

An inflection point occurs where the concavity changes, so here there is an inflection point at x=2. Plug 2 into the original equation to get the y value for the inflection point.