h(x)=x 2 βˆ’1h, left parenthesis, x, right parenthesis, equals, x, squared, minus, 1 Over which interval does h hh have a negative average rate of change? Choose 1 answer: Choose 1 answer: (Choice A) A βˆ’ 3 ≀ x ≀ 5 βˆ’3≀x≀5minus, 3, is less than or equal to, x, is less than or equal to, 5 (Choice B) B 1 ≀ x ≀ 4 1≀x≀41, is less than or equal to, x, is less than or equal to, 4 (Choice C) C βˆ’ 3 ≀ x ≀ 1 βˆ’3≀x≀1minus, 3, is less than or equal to, x, is less than or equal to, 1 (Choice D) D βˆ’ 1 ≀ x ≀ 5 βˆ’1≀x≀5minus, 1, is less than or equal to, x, is less than or equal to, 5 Show Calculator

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Answer:

(C)βˆ’3 ≀ x ≀ 1

Step-by-step explanation:

The average rate of change of function h over the interval [tex]a \leq x\leq b[/tex], is given by this expression:

[tex]\dfrac{h(b)-h(a)}{b-a}[/tex]

Given the function [tex]h(x)=x^2-1[/tex] on the interval:βˆ’ 3 ≀ x ≀ 1

[tex]h(1)=1^2-1=0\\h(-3)=(-3)^2-1=9-1=8[/tex]

The average rate of change:

[tex]\dfrac{h(b)-h(a)}{b-a}=\dfrac{0-8}{1-(-3)}=\dfrac{-8}{4}=-2[/tex]

Therefore, the function has a negative average rate of change over the interval βˆ’ 3 ≀ x ≀ 1.

CHECK:

(A)Average rate of change of h(x) over the interval βˆ’ 3 ≀ x ≀ 5=2

(B)Average rate of change of h(x) over the interval 1 ≀ x ≀ 4=5

(D)Average rate of change of h(x) over the interval βˆ’ 1 ≀ x ≀ 5=4