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Artyom0805 [142]
3 years ago
15

What is the average rate of change of h(x) over the interval -13

Mathematics
1 answer:
Ludmilka [50]3 years ago
5 0
The average rate of change of h(x) in the closed interval [ a, b ] is

Here [a, b ] = [ - 2, 2 ], thus
f(b) = f(2) = × 2³ - 2² = 1 - 4 = - 3
f(a) = × (- 2)³ - (- 2)² = - 1 - 4 = - 5
average rate of change = = =
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<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20x%20%7B%7D%5E%7B2%7D%20%20%2B%2012x%20%2B%2036" id="TexFormula1" title="f(x)
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Answer:

B

Step-by-step explanation:

We are given the function:

\displaystyle f(x) = x^2 + 12x + 36

And we want to determine the value of:

\displaystyle f^{-1}(225)

Let this value equal <em>a</em>. In other words:  

\displaystyle f^{-1}(225) = a

Then by the definition of inverse functions:

\displaystyle \text{If } f^{-1}(225) = a\text{, then } f(a) = 225

Hence:

\displaystyle f(a) =225 = (a)^2 + 12(a) + 36

Solve for <em>a: </em>

\displaystyle \begin{aligned} 225 &= a^2 + 12a + 36 \\ \\ a^2 + 12a -189 &= 0 \\ \\ (a + 21)(a-9) &= 0\end{aligned}

By the Zero Product Property:

\displaystyle a + 21 = 0 \text{ or } a - 9 = 0

Hence:

\displaystyle a = -21 \text{ or } a = 9

Thus, f(9) = 225. Consequently, f⁻¹(225) = 9.

In conclusion, our answer is B.

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