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Given the function h(x) = x2 + 6x + 5, determine the average rate of change of
the function over the interval-4 < x < -1.

Sagot :

The average rate of change of the function h(x) = x2 + 6x + 5 over the function -4 < x < 1 is 1

The given function is:

h(x)  = x² + 6x + 5

The average rate of a function h(x) over the interval a < x < b is given as:

[tex]\frac{dh}{dx}= \frac{h(b)-h(a)}{b-a}[/tex]

For the rate of change of h(x)  = x² + 6x + 5 in the interval -4 < x < -1.

a  =  -4,  b =  -1

[tex]h(-1) = (-1)^2 + 6(-1)+5\\h(-1) = 1 -6+5\\h(-1) = 0[/tex]

[tex]h(-4) = (-4)^2+6(-4)+5\\h(-4) = 16-24+5\\h(-4) = -3[/tex]

[tex]\frac{dh}{dx}= \frac{h(-1)-h(-4)}{-1-(-4)} \\\frac{dh}{dx}= \frac{0-(-3)}{3} \\\frac{dh}{dx}= \frac{3}{3} \\\frac{dh}{dx}=1[/tex]

The average rate of change of the function h(x) = x2 + 6x + 5 over the function -4 < x < 1 is 1

Learn more here: https://brainly.com/question/11883878

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