Question

a. find the open interval(s) on which the function is increasing and decreasing.
b. identify the functions local and absolute extreme values, if any, saying where they occur.
g(t) = -2t² + 3t - 1
values of t. in increasing order of t - value, the maximum values are g(□) = □, g(□) = □, and g(□) = □.
c. the function has a local maximum value at one value of t. the maximum value is g(□) = □.
d. there are no local maxima.

Explanation:

Step1: Find the derivative of \(g(t)\)

Differentiate \(g(t)=-2t^{2}+3t - 1\) using the power rule \((x^n)^\prime=nx^{n - 1}\).
\(g^\prime(t)=-4t + 3\)

Step2: Find the critical points

Set \(g^\prime(t)=0\), so \(-4t+3 = 0\).
Solve for \(t\): \(t=\frac{3}{4}\)

Step3: Determine the intervals of increase and decrease

• For \(t<\frac{3}{4}\), let \(t = 0\). Then \(g^\prime(0)=-4(0)+3=3>0\). So \(g(t)\) is increasing on the interval \((-\infty,\frac{3}{4})\)
• For \(t>\frac{3}{4}\), let \(t = 1\). Then \(g^\prime(1)=-4(1)+3=-1<0\). So \(g(t)\) is decreasing on the interval \((\frac{3}{4},\infty)\)

Step4: Find the local maximum

Since the function changes from increasing to decreasing at \(t=\frac{3}{4}\), we find \(g(\frac{3}{4})\)
\(g(\frac{3}{4})=-2(\frac{3}{4})^{2}+3(\frac{3}{4})-1=-2\times\frac{9}{16}+\frac{9}{4}-1=-\frac{9}{8}+\frac{18}{8}-\frac{8}{8}=\frac{-9 + 18-8}{8}=\frac{1}{8}\)

Answer:

C. The function has a local maximum value at one value of \(t\). The maximum value is \(g(\frac{3}{4})=\frac{1}{8}\)