aaliyahbrooks1865 aaliyahbrooks1865
  • 17-06-2022
  • Mathematics
contestada

h(x) = 1/8^3 -x^2 What is the average rate of change of h over the interval -2 ≤ x ≤ 2?​

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Medunno13
Medunno13 Medunno13
  • 30-06-2022

Answer: 0

Step-by-step explanation:

I will assume you meant [tex]h(x)=\frac{1}{8^{3}}-x^2[/tex], which is equal to [tex]h(x)=\frac{1}{512}-x^2[/tex].

[tex]f(-2)=\frac{1}{512}-(-2)^{2}=-\frac{2047}{512}\\\\f(2)=\frac{1}{512}-2^{2}=-\frac{2047}{512}\\\\\therefore \frac{f(2)-f(-2)}{2-(-2)}=\frac{-\frac{2047}{512}-\left(-\frac{2047}{512} \right)}{2-(-2)}=\boxed{0}[/tex]

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