Instantaneous Rate of Change (College Board AP® Calculus AB)

Revision Note

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The instantaneous rate of change is the slope of a graph at a specific point, rather than between two points
Consider the graph of a function $f$ with two points on the graph, $A$ and $B$

Using the labeling on the left,
- the average rate of change from $A$ to $B$ can be written as $\frac{f (x) - f (a)}{x - a}$
Using the labeling on the right,
- the average rate of change from $A$ to $B$ can be written as $\frac{f (a + h) - f (a)}{h}$
To find the instantaneous rate of change, consider finding the slope between $A$ and $B$ as point $B$ moves closer to point $A$
- As the distance between the two points becomes smaller, the slope will become a more accurate estimate for the instantaneous rate of change at $x = a$
- Therefore the instantaneous rate of change will be the limit as this distance tends to zero
The instantaneous rate of change at $x = a$ can be written as
- $\lim_{x \to a} \frac{f (x) - f (a)}{x - a}$ or
- $\lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$
These only give a valid answer if the relevant limit exists
They are both equivalent forms of the definition of the derivative of the function at $x = a$ , denoted by $f^{'} (a)$

A function $f (x)$ is defined by $f (x) = x^{2}$ .

Using the equation below, find the instantaneous rate of change of $f (x)$ at the point where $x = 2$ .

$f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

Answer:

Substitute $x = 2$ into the given formula

$f^{'} (2) = \lim_{h \to 0} \frac{f (2 + h) - f (2)}{h}$

Evaluate the function, $f (x) = x^{2}$ , at $x = 2 + h$ and $x = 2$

$f^{'} (2) = \lim_{h \to 0} \frac{{(2 + h)}^{2} - {(2)}^{2}}{h}$

Expand and simplify the numerator

$f^{'} (2) = \lim_{h \to 0} \frac{4 + 4 h + h^{2} - 4}{h} f^{'} (2) = \lim_{h \to 0} \frac{4 h + h^{2}}{h}$

Simplify the fraction by cancelling terms

$f^{'} (2) = \lim_{h \to 0} (4 + h)$

Evaluate the limit

$f^{'} (2) = 4$

the (exam) results speak for themselves:

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