Question

determine the intervals on which the graph of the function g(t)=3t^4 + 18t^3 - 7t - 7 is concave up and concave down. concave up: concave down:

Explanation:

Step1: Find the first - derivative

Differentiate $g(t)=3t^{4}+18t^{3}-7t - 7$ using the power rule. $g^\prime(t)=12t^{3}+54t^{2}-7$.

Step2: Find the second - derivative

Differentiate $g^\prime(t)$ using the power rule. $g^{\prime\prime}(t)=36t^{2}+108t$.

Step3: Set the second - derivative equal to zero

Solve $g^{\prime\prime}(t) = 0$. So, $36t^{2}+108t=36t(t + 3)=0$. The solutions are $t = 0$ and $t=-3$.

Step4: Test intervals

Choose test points in the intervals $(-\infty,-3)$, $(-3,0)$ and $(0,\infty)$.
For the interval $(-\infty,-3)$, let $t=-4$. Then $g^{\prime\prime}(-4)=36\times(-4)^{2}+108\times(-4)=576 - 432 = 144>0$.
For the interval $(-3,0)$, let $t=-1$. Then $g^{\prime\prime}(-1)=36\times(-1)^{2}+108\times(-1)=36 - 108=-72<0$.
For the interval $(0,\infty)$, let $t = 1$. Then $g^{\prime\prime}(1)=36\times1^{2}+108\times1=36 + 108 = 144>0$.

Answer:

Concave up: $(-\infty,-3),(0,\infty)$
Concave down: $(-3,0)$