Question

find the absolute maximum and minimum values of the function over the indicated interval.

$f(x)=4x^{2}+8$
(a) $4,5$
(b) $-5,5$

(a) the absolute maximum value is $\square$ at $x=\square$
(use a comma to separate answers as needed.)

the absolute minimum value is $\square$ at $x=\square$
(use a comma to separate answers as needed.)

(b) the absolute maximum value is $\square$ at $x=\square$
(use a comma to separate answers as needed.)

the absolute minimum value is $\square$ at $x=\square$
(use a comma to separate answers as needed.)

Explanation:

Step1: Analyze the function \(f(x) = 4x^{2}+8\)

The function \(y = ax^{2}+bx + c\) (\(a = 4\), \(b=0\), \(c = 8\)) is a parabola. Since \(a=4>0\), the parabola opens upward. The vertex form of a parabola is \(y=a(x - h)^{2}+k\). For \(y = 4x^{2}+8\), \(h = 0\) and \(k = 8\). The vertex is \((0,8)\). The derivative \(f^{\prime}(x)=8x\). Set \(f^{\prime}(x)=0\), we get \(x = 0\).

Step2: Solve for interval \([4,5]\)

Evaluate \(f(x)\) at the endpoints of the interval.
When \(x = 4\), \(f(4)=4\times4^{2}+8=4\times16 + 8=64 + 8=72\).
When \(x = 5\), \(f(5)=4\times5^{2}+8=4\times25+8=100 + 8=108\).

Step3: Solve for interval \([-5,5]\)

Evaluate \(f(x)\) at the endpoints \(x=-5\) and \(x = 5\) and the critical point \(x = 0\).
When \(x=-5\), \(f(-5)=4\times(-5)^{2}+8=4\times25 + 8=100+8 = 108\).
When \(x = 0\), \(f(0)=4\times0^{2}+8=8\).
When \(x = 5\), \(f(5)=4\times5^{2}+8=108\).

Answer:

(A) The absolute maximum value is \(108\) at \(x = 5\). The absolute minimum value is \(72\) at \(x = 4\).
(B) The absolute maximum value is \(108\) at \(x=-5,5\). The absolute minimum value is \(8\) at \(x = 0\).