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.

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step-by-step explanation:

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.