f(x)=x^{2}+10\nover which interval does f have a positive average rate of change?\nchoose 1 answer:\na…

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 given by (\frac{f(b)-f(a)}{b - a}).

Step2: Calculate for option A ([-3,1])

First, find (f(-3)) and (f(1)). (f(-3)=(-3)^{2}+10=9 + 10=19), (f(1)=1^{2}+10=1+10 = 11). The average rate of change is (\frac{f(1)-f(-3)}{1-(-3)}=\frac{11 - 19}{1 + 3}=\frac{-8}{4}=-2).

Step3: Calculate for option B ([-1,2])

Find (f(-1)) and (f(2)). (f(-1)=(-1)^{2}+10=1+10 = 11), (f(2)=2^{2}+10=4 + 10=14). The average rate of change is (\frac{f(2)-f(-1)}{2-(-1)}=\frac{14 - 11}{2+1}=\frac{3}{3}=1).

Step4: Calculate for option C ([-4,-1])

Find (f(-4)) and (f(-1)). (f(-4)=(-4)^{2}+10=16+10 = 26), (f(-1)=(-1)^{2}+10=1+10 = 11). The average rate of change is (\frac{f(-1)-f(-4)}{-1-(-4)}=\frac{11 - 26}{-1 + 4}=\frac{-15}{3}=-5).

Step5: Calculate for option D ([-3,3])

Find (f(-3)) and (f(3)). (f(-3)=(-3)^{2}+10=9+10 = 19), (f(3)=3^{2}+10=9 + 10=19). The average rate of change is (\frac{f(3)-f(-3)}{3-(-3)}=\frac{19 - 19}{3+3}=0).

Answer:

B. ([-1,2])