We are given the function $f(x) = x^2 - 2$ and need to find the average rate of change (ARoC) over different intervals. The general formula for the average rate of change of a function $f(x)$ from $x = a$ to $x = b$ is:

$ARoC = \frac{f(b) - f(a)}{b - a}$

Part (a) Find the ARoC from 1 to $1 + h$

The interval is from $x = 1$ to $x = 1 + h$. To calculate this, we need to:

1. Evaluate $f(1)$.
2. Evaluate $f(1 + h)$.
3. Use the formula for ARoC.

Let's proceed step-by-step:

1. Evaluate $f(1)$: $f(1) = 1^2 - 2 = 1 - 2 = -1$

2. Evaluate $f(1 + h)$: $f(1 + h) = (1 + h)^2 - 2 = 1 + 2h + h^2 - 2 = h^2 + 2h - 1$

3. Apply the ARoC formula: $ARoC = \frac{f(1 + h) - f(1)}{(1 + h) - 1} = \frac{(h^2 + 2h - 1) - (-1)}{h} = \frac{h^2 + 2h}{h} = h + 2$

So, the average rate of change from $1$ to $1 + h$ is:

$ARoC = h + 2$

Part (b) Find the ARoC from 1 to 276 (using part a)

Now, we need to find the ARoC from $x = 1$ to $x = 276$. We can apply the result from part (a), where we use the formula for the ARoC from 1 to $1 + h$, but instead substitute $h = 276 - 1 = 275$:

$ARoC = 275 + 2 = 277$

So, the average rate of change from $1$ to $276$ is:

$ARoC = 277$

Part (c) Find the ARoC from 1 to 1.36 (using part a)

Lastly, we need to find the ARoC from $x = 1$ to $x = 1.36$. Here, we can again use the formula from part (a), where $h = 1.36 - 1 = 0.36$:

$ARoC = 0.36 + 2 = 2.36$

So, the average rate of change from $1$ to $1.36$ is:

$ARoC = 2.36$

Summary of Results:

• (a) The ARoC from 1 to $1 + h$ is $h + 2$.
• (b) The ARoC from 1 to 276 is $277$.
• (c) The ARoC from 1 to 1.36 is $2.36$.

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Would you like a more detailed explanation on any of these steps?

Here are some related questions to explore further:

1. How would you derive the formula for average rate of change from the definition of a derivative?
2. How does the concept of ARoC relate to the instantaneous rate of change at a specific point?
3. What is the geometric interpretation of the average rate of change for a quadratic function?
4. How would the ARoC change if the function was $f(x) = x^3 - 2x$?
5. How would you compute the ARoC for an interval where one of the endpoints is negative?

Tip: To simplify the evaluation of ARoC, it's helpful to use the fact that the function $f(x) = x^2 - 2$ is a polynomial, making it easy to apply basic algebraic operations on the expressions.