Maths help?

f(x) = 3x³ - 4x + 1
f'(x) = 9x² - 4
f"(x) = 18x

Consider f'(x) = 0
9x² - 4 = 0
(3x + 2)(3x - 2) = 0
x = -2/3 or x = 2/3

When x = -2/3:
f(x) = 3(-2/3)³ - 4(-2/3) + 1 = 25/9
f"(x) = 18(-2/3) < 0
Hence, local maximum at (-2/3, 25/9)

When x = 2/3:
f(x) = 3(2/3)³ - 4(2/3) + 1 = -7/9
f"(x) = 18(2/3) > 0
Hence, local minimum at (2/3, -7/9)

f(x) = 3x³ - 4x + 1 ← this is a function

f'(x) = 9x² - 4 ← this is its derivative


You can obtain a stationary point when the derivative is zero.

f'(x) = 0

9x² - 4 = 0

9x² = 4

x² = 4/9

x = ± 2/3


x_______-∞____- 2/3____2/3____+∞

(9x² - 4)_____+___0___-__0___+

f'(x)________+____0___-__0___+

f(x)________↑____↔__↓__↔___↑


f(x) = 3x³ - 4x + 1 → when: x = - 2/3

f(- 2/3) = 3.(- 2/3)³ - 4.(- 2/3) + 1

f(- 2/3) = 25/9 → first stationary point @ A (- 2/3 ; 25/9)


f(x) = 3x³ - 4x + 1 → when: x = 2/3

f(2/3) = 3.(2/3)³ - 4.(2/3) + 1

f(2/3) = - 7/9 → second stationary point @ B (2/3 ; - 7/9)

is there an exponent?  if there is you should write it as 3x^3

I'm assuming it's 3x^3 - 4x + 1

Stationary point means derivative is 0, so find the derivative

Now, test points around where the derivative is 0. The smallest of the points is -2/3 so test a point less than -2/3 and then greater than -2/3, but this second one also has to be less than the other stationary point. 

 if the two test points have the same sign when you plug them into f'(x), the derivative, it is not a max or min.  if it goes from negative to positive you have a min and if it goes from positive to negative you have a max.

Let me know if you need any specific help with any of these parts.