To solve this problem, we follow these steps:

Step 1: Calculate the average rate of change over the interval $[0, 2]$.

The formula for the average rate of change of a function $f(x)$ over an interval $[a, b]$ is:

$\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}$

Here, $f(x) = x^2 + 2$, $a = 0$, and $b = 2$. Compute $f(0)$ and $f(2)$:

$f(0) = 0^2 + 2 = 2$ $f(2) = 2^2 + 2 = 6$

The average rate of change is: $\text{Average rate of change} = \frac{f(2) - f(0)}{2 - 0} = \frac{6 - 2}{2} = 2$

Step 2: Find the instantaneous rate of change of $f(x)$.

The instantaneous rate of change of $f(x)$ at any point $x$ is given by its derivative $f'(x)$. Differentiate $f(x)$:

$f'(x) = 2x$

Step 3: Set the instantaneous rate of change equal to the average rate of change.

We are tasked with finding the value of $x$ where $f'(x) = 2$. Set up the equation:

$2x = 2$

Solve for $x$: $x = 1$

Step 4: Verify the result.

At $x = 1$, the instantaneous rate of change is: $f'(1) = 2(1) = 2$ which matches the average rate of change over $[0, 2]$.

Final Answer:

The time at which the instantaneous rate of change equals the average rate of change is: $x = 1$

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Related Questions:

1. What is the significance of the average and instantaneous rate of change in real-world scenarios?
2. How would the solution change if the interval was $[1, 3]$?
3. What is the geometric interpretation of the solution in terms of the tangent and secant lines?
4. Can this technique be applied to functions other than polynomials? If so, how?
5. How do the rates of change vary if the function is $f(x) = x^3 + 2$?

Tip:

For any continuous and differentiable function, the Mean Value Theorem guarantees at least one point where the instantaneous rate of change equals the average rate of change over an interval.