Question

What are the points of inflection of `f(x)=x^3-2x^2 - x`?

Answer 1

`(2/3 , -34/27 )`

Explanation:

To determine points of inflection , require to find the 2nd derivative,equate it to zero and solve.

f'(x) `= 3x^2 - 4x - 1`

f''(x) = 6x - 4

To find a possible point of inflection, set f''(x) = 0

6x - 4 = 0 → `x = 2/3`

Require to test for a change in sign either side of this value.

choose, say x = 0 and evaluate f''(0)

f''(0) = 0 - 4 = -4 , hence function is concave down to the left

now choose say x = 1 and evaluate f''(1)

f''(1) = 6 - 4 = 2 , hence function is concave up to the right

and `f(2/3) = 3(2/3)^2 - 4(2/3) -2/3 = -34/27`

Conclusion: The function changes from concave down to concave up at `(2/3 , -34/7) " hence point of inflection "`
graph{x^3-2x^2-x [-10, 10, -5, 5]}