Sure, let's solve each part of the question step-by-step.
### Part (a): Average Rate of Change
The average rate of change of a function [tex]\( y = f(x) \)[/tex] over an interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[
\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}
\][/tex]
Here, we are given the function [tex]\( y = f(x) = x^1 - 9 \)[/tex] and the interval [tex]\([2, 6]\)[/tex]. So, let's calculate the values of [tex]\( f(2) \)[/tex] and [tex]\( f(6) \)[/tex]:
1. Calculate [tex]\( f(2) \)[/tex]:
[tex]\[
f(2) = 2^1 - 9 = 2 - 9 = -7
\][/tex]
2. Calculate [tex]\( f(6) \)[/tex]:
[tex]\[
f(6) = 6^1 - 9 = 6 - 9 = -3
\][/tex]
Next, we use these values to find the average rate of change:
[tex]\[
\text{Average rate of change} = \frac{f(6) - f(2)}{6 - 2} = \frac{-3 - (-7)}{6 - 2} = \frac{-3 + 7}{6 - 2} = \frac{4}{4} = 1
\][/tex]
So, the average rate of change of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] over the interval [tex]\([2, 6]\)[/tex] is [tex]\( 1.0 \)[/tex].
### Part (b): Instantaneous Rate of Change
The instantaneous rate of change at a point [tex]\( x \)[/tex] is given by the derivative of the function [tex]\( y = f(x) \)[/tex] evaluated at that point.
The function given is:
[tex]\[
y = f(x) = x^1 - 9
\][/tex]
To find the instantaneous rate of change at [tex]\( x = 4 \)[/tex], we first find the derivative of [tex]\( f(x) \)[/tex]. Since [tex]\( f(x) = x - 9 \)[/tex], the derivative [tex]\( f'(x) \)[/tex] is:
[tex]\[
f'(x) = \frac{d}{dx} (x - 9) = 1
\][/tex]
Now, we evaluate the derivative at [tex]\( x = 4 \)[/tex]:
[tex]\[
f'(4) = 1
\][/tex]
So, the instantaneous rate of change of [tex]\( y \)[/tex] at [tex]\( x = 4 \)[/tex] is [tex]\( 1 \)[/tex].
### Summary
- The average rate of change of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] over the interval [tex]\([2, 6]\)[/tex] is [tex]\( 1.0 \)[/tex].
- The instantaneous rate of change of [tex]\( y \)[/tex] at [tex]\( x = 4 \)[/tex] is [tex]\( 1 \)[/tex].
