find the absolute maximum and minimum values of the function over the indicated interval. f(x)=2x² + 5 (a)…

Explanation:

Step1: Analyze the function

The function $f(x)=2x^{2}+5$ is a parabola opening upwards ($a = 2>0$ in $y = ax^{2}+bx + c$, here $b = 0,c = 5$), and its vertex - form is $y=2(x - 0)^{2}+5$, so the vertex is at $(0,5)$.

Step2: Evaluate the function at endpoints and critical points for interval [-6,6]

The critical point of $f(x)$ is found by taking the derivative $f^\prime(x)=4x$. Set $f^\prime(x)=0$, we get $4x = 0$, so $x = 0$. Evaluate $f(x)$ at the endpoints $x=-6,x = 6$ and the critical - point $x = 0$: When $x=-6$, $f(-6)=2\times(-6)^{2}+5=2\times36 + 5=72 + 5=77$. When $x = 6$, $f(6)=2\times6^{2}+5=2\times36+5=72 + 5=77$. When $x = 0$, $f(0)=2\times0^{2}+5=5$.

Answer:

The absolute minimum value is 5 at x = 0.