a youtuber tracks the growth of subscribers (in hundreds) t weeks after the first video. a model…

Explanation:

Step1: Find the first derivative

Use the sum - rule and basic derivative formulas. The derivative of (y = 9\ln(t + 1)) is (y^\prime=\frac{9}{t + 1}), the derivative of (y=-2t) is (y^\prime=-2), and the derivative of (y=\frac{1}{2}t^{2}) is (y^\prime=t). So, (S^\prime(t)=\frac{9}{t + 1}-2 + t).

Step2: Find the second derivative

Use the quotient - rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}) for (\frac{9}{t + 1}) (where (u = 9), (v=t + 1), (u^\prime=0), (v^\prime = 1)). (S^{\prime\prime}(t)=-\frac{9}{(t + 1)^{2}}+1).

Step3: Solve (S^{\prime\prime}(t)=0) for inflection points

Set (S^{\prime\prime}(t)=0), then (-\frac{9}{(t + 1)^{2}}+1 = 0). Rearrange to get (\frac{9}{(t + 1)^{2}}=1), then ((t + 1)^{2}=9). Take square roots: (t+1=\pm3). Since (t\geq0), (t + 1=3), so (t = 2).

Step4: Test intervals for concavity

• For (a) (concave down): Test an interval (t\in[0,2)). Let (t = 1), (S^{\prime\prime}(1)=-\frac{9}{(1 + 1)^{2}}+1=-\frac{9}{4}+1=-\frac{5}{4}<0). So (S(t)) is concave down on ([0,2)).
• For (b) (concave up): Test an interval (t\in(2,\infty)). Let (t=3), (S^{\prime\prime}(3)=-\frac{9}{(3 + 1)^{2}}+1=-\frac{9}{16}+1=\frac{7}{16}>0). So (S(t)) is concave up on ((2,\infty)).
• For (c) (inflection point): Since (S^{\prime\prime}(t)) changes sign at (t = 2), the inflection point is at (t = 2).

Answer:

a. (S(t)) is concave down on ([0,2)). b. (S(t)) is concave up on ((2,\infty)). c. The inflection point is at (t = 2).