find the absolute maximum and minimum values of the function over the indicated interval.\n\n$f(x)=4x^{2}+8$\…

find the absolute maximum and minimum values of the function over the indicated interval.\n\n$f(x)=4x^{2}+8$\n(a) $4,5$\n(b) $-5,5$\n\n(a) the absolute maximum value is $\\square$ at $x=\\square$\n(use a comma to separate answers as needed.)\n\nthe absolute minimum value is $\\square$ at $x=\\square$\n(use a comma to separate answers as needed.)\n\n(b) the absolute maximum value is $\\square$ at $x=\\square$\n(use a comma to separate answers as needed.)\n\nthe absolute minimum value is $\\square$ at $x=\\square$\n(use a comma to separate answers as needed.)

find the absolute maximum and minimum values of the function over the indicated interval.\n\n$f(x)=4x^{2}+8$\n(a) $4,5$\n(b) $-5,5$\n\n(a) the absolute maximum value is $\\square$ at $x=\\square$\n(use a comma to separate answers as needed.)\n\nthe absolute minimum value is $\\square$ at $x=\\square$\n(use a comma to separate answers as needed.)\n\n(b) the absolute maximum value is $\\square$ at $x=\\square$\n(use a comma to separate answers as needed.)\n\nthe absolute minimum value is $\\square$ at $x=\\square$\n(use a comma to separate answers as needed.)

Answer

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).