a start - up sells a seasonal gadget. its monthly revenue (in thousands of dollars) since launch is modeled…

a start - up sells a seasonal gadget. its monthly revenue (in thousands of dollars) since launch is modeled by (r(t)=-0.7t^{3}+8t^{2}+50t + 240), (tgeq0). here (t) is the number of months since launch. a. find the interval of time where (r) is concave up. (round to 1 decimal place as needed) (use interval notation) b. find the interval of time where (r) is concave down. c. find any inflection point(s).

a start - up sells a seasonal gadget. its monthly revenue (in thousands of dollars) since launch is modeled by (r(t)=-0.7t^{3}+8t^{2}+50t + 240), (tgeq0). here (t) is the number of months since launch. a. find the interval of time where (r) is concave up. (round to 1 decimal place as needed) (use interval notation) b. find the interval of time where (r) is concave down. c. find any inflection point(s).

Answer

Explanation:

Step1: Find the first - derivative

The revenue function is $R(t)=- 0.7t^{3}+8t^{2}+50t + 240$, $t\geq0$. Using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $R^\prime(t)=-2.1t^{2}+16t + 50$.

Step2: Find the second - derivative

Differentiate $R^\prime(t)$ with respect to $t$. Using the power rule again, $R^{\prime\prime}(t)=-4.2t + 16$.

Step3: Find where $R^{\prime\prime}(t)=0$ (for inflection points)

Set $R^{\prime\prime}(t)=0$. So, $-4.2t + 16 = 0$. Solving for $t$ gives $t=\frac{16}{4.2}\approx3.8$.

Step4: Test intervals for concavity

We have two intervals to test: $[0,3.8)$ and $(3.8,\infty)$. For $t = 0$ (in the interval $[0,3.8)$), $R^{\prime\prime}(0)=16>0$, so $R(t)$ is concave up on the interval $[0,3.8)$. For $t = 4$ (in the interval $(3.8,\infty)$), $R^{\prime\prime}(4)=-4.2\times4 + 16=-0.8<0$, so $R(t)$ is concave down on the interval $(3.8,\infty)$.

a. # Answer: $[0,3.8)$ b. # Answer: $(3.8,\infty)$ c. # Answer: $t\approx3.8$