Answer:
A) g is increasing, and the graph of g is concave up.
Step-by-step explanation:
g'(x) = ∫₀ˣ e^(-t³) dt
Since e^(-t³) is always positive, ∫₀ˣ e^(-t³) dt is positive when x > 0. So the function is increasing.
Find g"(x) by taking the derivative using second fundamental theorem of calculus:
g"(x) = e^(-x³)
g"(x) is always positive, so the function is always concave up.
The answer is 0.0125 or 1/80. Hope that helps! :)
Answer:
The answer is on the attachment.
