f(x) = 2x^3 - 3x^2 - 72x + 7
f'(x) = 6x^2 - 6x - 72 = 6(x^2-x-12)
f"(x) = 12x-6 = 6(2x-1)
f'=0 at x= -3 and 4
f"=0 at x = 1/2
Now it's easy to answer the questions.
f is increasing where f' > 0
f is concave up where f" > 0
no asymptotes, so critical points where f' = 0
Consider the function
f (x) = 2x^3 - 3x^2 - 72x + 7
Find and classify the critical points. Identify the intervals of increase and decrease, and state the intervals of concavity.
1 answer
1 answer