Hello there!

can i get some with this calculus problem? don't forget to show your work or any useful information i should be aware of.

a. increasing: (-∞, 0) ∪ (0, 5); decreasing: (5, ∞)

b. concave up: (0, 3 1/3); concave down: (-∞, 0) ∪ (3 1/3, ∞)

c. relative extrema: x=5; inflection points: x=0, x=3 1/3

d. see attached. The green curve has the shape of f(x), but is scaled by 1/3 so it will fit on the same graph with f'(x). So, f(x) will be scaled vertically by a factor of 3 from that, and may have any vertical offset.

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Step-by-step explanation:

a. f is increasing where f' is positive: everywhere to the left of x=5, except at x=0, where f'(0) = 0. f is decreasing where f' is negative, to the right of x=5.

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b. f is concave upward where f'' is positive, which is where f' is increasing. That interval is between x=0 and x=3 1/3. f is concave downward where f' is decreasing, on intervals outside [0, 3 1/3].

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c. The extreme points of f are where f' is zero and f'' is not zero. The only one is at x=5. The points of inflection are where f'' is zero, at the extremes of f'(x): x=0, x=3 1/3. They are also at the places where concavity changes as described in b.

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d. The function f is the integral of the first derivative, plus some arbitrary constant. We have shown in the attachment a function that could be 1/3·f(x) (green curve). The vertical scaling is so it will fit on the graph.