How maximum f (x)=x^3 - 12x + 2

Question

aum · Answer

Find the derivative, equate it to zero and solve for x.
Find the second derivative and test each critical point found in step one.  For maximum, f''(c) < 0.

anonymous · Answer

I don't understand
Please Explain to me

aum · Answer

Do you know how to take the derivative?
If yes, can you find the derivative of f(x)?

anonymous · Answer

No l can't

anonymous · Answer

Please halp me

aum · Answer

$$
 f (x)=x^3 - 12x + 2 \
f'(x) = 3x^2 - 12 = 0 \
3x^2 = 12 \
x^2 = 4 \
x = -2 \text{ or } x = 2  \
f''(x) = 6x \
f''(-2) = -12 \
\text{Since f''(-2) < 0 x = -2 gives  a maximum.}\
\text{The maximum value is } f(-2) = (-2)^3 - 12(-2) + 2 = -8 + 24 + 2 = 18
$$

anonymous · Answer

Thank you
Can you answer to another question?

anonymous · Answer

X^3-x^2+1

aum · Answer

Follow the method shown above and apply the same method for the second problem.

anonymous · Answer

I could not be resolved
Please Explain it