find the absolute maximum value and the absolute minimum value, if any, of the function. (if an answer does…

Explanation:

Step1: Find the derivative

The derivative of $g(x)=-x^{2}+2x + 7$ using the power - rule $(x^n)'=nx^{n - 1}$ is $g'(x)=-2x + 2$.

Step2: Find critical points

Set $g'(x)=0$. So, $-2x + 2 = 0$. Solving for $x$ gives $2x=2$, then $x = 1$.

Step3: Determine the second - derivative

The second - derivative $g''(x)=-2<0$. Since $g''(1)=-2<0$, the function has a maximum at $x = 1$.

Step4: Calculate the maximum value

Substitute $x = 1$ into $g(x)$: $g(1)=-(1)^{2}+2(1)+7=-1 + 2+7=8$.

Step5: Analyze the behavior of the function

Since $g(x)=-x^{2}+2x + 7$ is a quadratic function with $a=-1<0$, the parabola opens downwards. As $x\to\pm\infty$, $g(x)\to-\infty$. So, there is no absolute minimum.

Answer:

maximum: 8 minimum: DNE