a. find the open intervals on which the function is increasing and those on which it is decreasing. b…

a. find the open intervals on which the function is increasing and those on which it is decreasing. b. identify the functions local extreme values, if any, saying where they occur. g(t)= - 4t² - 3t + 3 b. find each local maximum, if any. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. (simplify your answers.) a. the function has a local maximum value at one value of t. the maximum value is g( ) = b. the function has a local maximum value at two values of t. in increasing order of t - value, the maximum values are g( ) = and g( ) = c. the function has a local maximum value at three values of t. in increasing order of t - value, the maximum values are g( ) =, g( ) =, and g( ) = d. there are no local maxima. find each local minimum, if any. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. (simplify your answers.) a. the function has a local minimum value at one value of t. the minimum value is g( ) =

a. find the open intervals on which the function is increasing and those on which it is decreasing. b. identify the functions local extreme values, if any, saying where they occur. g(t)= - 4t² - 3t + 3 b. find each local maximum, if any. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. (simplify your answers.) a. the function has a local maximum value at one value of t. the maximum value is g( ) = b. the function has a local maximum value at two values of t. in increasing order of t - value, the maximum values are g( ) = and g( ) = c. the function has a local maximum value at three values of t. in increasing order of t - value, the maximum values are g( ) =, g( ) =, and g( ) = d. there are no local maxima. find each local minimum, if any. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. (simplify your answers.) a. the function has a local minimum value at one value of t. the minimum value is g( ) =

Answer

Explanation:

Step1: Find the derivative

The derivative of $g(t)=-4t^{2}-3t + 3$ using the power - rule $(x^n)'=nx^{n - 1}$ is $g'(t)=-8t-3$.

Step2: Find the critical points

Set $g'(t) = 0$. So, $-8t-3=0$. Solving for $t$ gives $t=-\frac{3}{8}$.

Step3: Determine the intervals of increase and decrease

Choose a test - point in the interval $(-\infty,-\frac{3}{8})$, say $t=-1$. Then $g'(-1)=-8\times(-1)-3 = 5>0$, so the function is increasing on the interval $(-\infty,-\frac{3}{8})$. Choose a test - point in the interval $(-\frac{3}{8},\infty)$, say $t = 0$. Then $g'(0)=-8\times0 - 3=-3<0$, so the function is decreasing on the interval $(-\frac{3}{8},\infty)$.

Step4: Find local extreme values

Since the function changes from increasing to decreasing at $t = -\frac{3}{8}$, there is a local maximum at $t=-\frac{3}{8}$. The local maximum value is $g(-\frac{3}{8})=-4\times(-\frac{3}{8})^{2}-3\times(-\frac{3}{8})+3=-4\times\frac{9}{64}+\frac{9}{8}+3=-\frac{9}{16}+\frac{18}{16}+3=\frac{-9 + 18}{16}+3=\frac{9}{16}+3=\frac{9+48}{16}=\frac{57}{16}$. There are no local minimums.

Answer:

a. The function is increasing on the interval $(-\infty,-\frac{3}{8})$ and decreasing on the interval $(-\frac{3}{8},\infty)$. b. For local maximum: A. The function has a local maximum value at one value of $t$. The maximum value is $g(-\frac{3}{8})=\frac{57}{16}$. For local minimum: D. There are no local minimums.