Answer:
The average rate of change of [tex]f(x) = x^2 +6x+13[/tex] is 12
The average rate of change of [tex]f(x) =- x^2 +6x+13[/tex] is 10
Step-by-step explanation:
The average rate of change of f(x) over an interval between 2 points (a ,f(a)) and (b ,f(b)) is the slope of the secant line connecting the 2 points.
We can calculate the average rate of change between the 2 points by
[tex]\frac{f(b) - f(a)}{b -a}[/tex]-------------------(1)
(1) The average rate of change of the function [tex]f(x) = x^2 +6x+13[/tex] over the interval 1 ≤ x ≤ 5
f(a) = f(1)
[tex]f(1) = (1)^2 +6(1) + 13[/tex]
f(1) =1+6+13
f(a) = 20---------------------(2)
f(b) = f(5)
[tex]f(5) = (5)^2 +6(5)+13[/tex]
f(5) = 25 +30 +13
f(5) = 68-----------------------(3)
The average rate of change between (1 ,20) and (5 ,68 ) is
Substituting eq(2) and(3) in (1)
=[tex]\frac{f(5) - f(1)}{5-1}[/tex]
=[tex]\frac{68 -20}{5-1}[/tex]
= [tex]\frac{48}{4}[/tex]
=12
This means that the average of all the slopes of lines tangent to the graph of f(x) between (1 ,20) and (5 ,68 ) is 12
(2) The average rate of change of the function [tex]f(x) = -x^2 +6x+13[/tex] over the interval -1 ≤ x ≤ 5
f(a) = f(-1)
[tex]f(1) = (-1)^2 +6(-1) + 13[/tex]
f(1) =1-6+13
f(1) = 8---------------------(4)
f(b) = f(5)
[tex]f(5) = (5)^2 +6(5)+13[/tex]
f(5) = 25 +30 +13
f(5) = 68-----------------------(5)
The average rate of change between (-1 ,8) and (5 ,68 ) is
Equation (1) becomes
[tex]\frac{f(5) - f(-1)}{5-(-1)}[/tex]
On substituting the values
=[tex]\frac{68 - 8}{5-(-1)}[/tex]
=[tex]\frac{60}{5+1}[/tex]
=[tex]\frac{60}{6}[/tex]
= 10
This means that the average of all the slopes of lines tangent to the graph of f(x) between (-1 ,8) and (5 ,68 ) is 10
