f(x) = x^3 − 3x + 2
On differentiating, we get
f'(x) = 3x^2 − 3
For critical points, put f'(x) = 0
then I =
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f(x) = x^3 − 3x + 2
On differentiating, we get
f'(x) = 3x^2 − 3
For critical points, put f'(x) = 0
then I =
MathBot Answer:
MathBot is working on a solution to your problem.