Let f(x)= x^3-9x^2 +9x-5. Find the open intervals on which f is concave up (down). Then determine the x-coordi

let f(x)= x^3-9x^2 +9x-5. Find the open intervals on which f is concave up (down). Then determine the x-coordinates of all inflection points of f.

1.  
f is concave up on the intervals

2.  
f is concave down on the intervals

3.  
The inflection points occur at x =

f(x) = x^3 - 9x^2 + 9x - 5

When we see the word concavity, we know to look for the second derivative.

f'(x) = 3x^2 - 18x + 9

f''(x) = 6x - 18

We need to find where the second derivative changes sign, because this will tell us where our points of inflection are.

As such, we set f''(x) = 0

f''(x) = 0
6x-18 = 0
6x = 18
x = 3

Now, we test to the left and right of x = 3 to see where f''(x) is positive and where it is negative, as it is 0 at x = 3.

Picking random points we find:

f''(0) = 6(0) - 18 = -18
f'(5) = 6(5) - 18 = 30 - 18 = 12

Clearly, f''(x) is negative to the left or x = 3, and positive to the right of x = 3. We just simply need to know the f''(x) > 0 implies upward concavity while f''(x) < 0 implies downward concavity.

Answers:

1) (3, ∞)

2) (-∞, 3)

3) x = 3