select the correct answer. consider this function. f(x)=6log₂x - 3 over which interval is function f…

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function $y = f(x)$ over the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$.

Step2: Calculate the rate - of - change for option A

For $f(x)=6\log_2x - 3$ and the interval $[1,2]$: $f(1)=6\log_21-3=6\times0 - 3=-3$ $f(2)=6\log_22-3=6\times1 - 3 = 3$ The average rate of change is $\frac{f(2)-f(1)}{2 - 1}=\frac{3-(-3)}{1}=6$.

Step3: Calculate the rate - of - change for option B

For the interval $[2,6]$: $f(2)=3$ $f(6)=6\log_26-3=6\frac{\ln6}{\ln2}-3\approx6\times2.585 - 3=15.51 - 3 = 12.51$ The average rate of change is $\frac{f(6)-f(2)}{6 - 2}=\frac{12.51 - 3}{4}=\frac{9.51}{4}=2.3775$.

Step4: Calculate the rate - of change for option C

For the interval $[\frac{1}{2},1]$: $f(\frac{1}{2})=6\log_2\frac{1}{2}-3=6\times(- 1)-3=-6 - 3=-9$ $f(1)=-3$ The average rate of change is $\frac{f(1)-f(\frac{1}{2})}{1-\frac{1}{2}}=\frac{-3-(-9)}{\frac{1}{2}}=\frac{6}{\frac{1}{2}} = 12$.

Step5: Calculate the rate - of change for option D

For the interval $[\frac{1}{8},\frac{1}{2}]$: $f(\frac{1}{8})=6\log_2\frac{1}{8}-3=6\times(-3)-3=-18 - 3=-21$ $f(\frac{1}{2})=-9$ The average rate of change is $\frac{f(\frac{1}{2})-f(\frac{1}{8})}{\frac{1}{2}-\frac{1}{8}}=\frac{-9-(-21)}{\frac{3}{8}}=\frac{12}{\frac{3}{8}}=32$.

Answer:

D. $[\frac{1}{8},\frac{1}{2}]$