The range of the function f (x) = $\sqrt { 20 + 8 x - x ^ { 2 } }$ is a closed interval [a; b]. Find its length b - a.
A)1
B)2
C)3
D)4
E)5
F)6
G)7
H)9

Question

Quizplus · Accepted Answer

The function $ f(x) = \sqrt{20 + 8x - x^2} $ is defined where the expression under the square root is non-negative. The quadratic $ 20 + 8x - x^2 $ can be rewritten as $ -(x^2 - 8x + 20) $. Completing the square for $ x^2 - 8x + 20 $ gives $ (x-4)^2 + 4 $. Thus, the maximum value of $ 20 + 8x - x^2 $ is 24, and the minimum is 0. Therefore, the range of $ f(x) $ is $[0, \sqrt{24}]$, and the length of this interval is $\sqrt{24} - 0 = 2\sqrt{6}$. Since $\sqrt{24} = 2\sqrt{6}$, the length of the interval is 6.

The Range of the Function F (X) = 20+8x−x2\sqrt { 20 + 8 x - x ^ { 2 } }20+8x−x2​

The Range of the Function F (X) = $\sqrt { 20 + 8 x - x ^ { 2 } }$