Help with local minimum

Let f(x) = e^(x³ - 2x² - 4x + 5). Then f has a local minimum at x = ?
(I'm thinking 2 for this one, but I'm not sure if it's -2)

f '(x) tells you the slope at any point (x,y) along the graph of f(x). Notice that a min/max usually occurs where the slope is 0 or where a horizontal tangent occurs. Therefore, the first step is to set f '(x) = 0.

f '(x) = (3x^2 - 4x - 4)*e^(x^3 - 2x² - 4x + 5)

Note that e^(x^3 - 2x² - 4x + 5) can never be 0, so:

3x² - 4x - 4 = 0

3x² - 6x + 2x - 4 = 0

3x(x - 2) + 2(x - 2) = 0

(3x + 2)(x - 2) = 0

x = -2/3, 2

Now to perform the first derivative test to check for minimum:

x = -2/3:

Point to the left: -0.7, Sign: Negative
Point to the right: -0.65, Sign: Negative

Neither a min nor a max.

x = 2:

Point the left: 1.5, Sign: Negative
Point to the right: 2.5, Sign: Positive

Minimum occurs at x = 2.

f(x) = e^(x^3 - 2x^2 - 4x + 5)

f'(x) = e^(x^3 - 2x^2 - 4x + 5) • (3x^2 - 4x - 4) = 0

3x^2 - 4x - 4 = 0
(3x + 2)(x - 2) = 0
3x + 2 = 0 or x - 2 = 0
3x = -2.............x = 2
x = -2/3

The local minimum occurs at x = 2.

f'(x) = (3x^2 - 4x - 4)*e^(x^3 - 2x^2 - 4x + 5)
3x^2 - 4x - 4 = 0
(3x + 2)(x - 2) = 0
The local minimum is x=2.