A ball's height is $h(t) = -4.9t^2 + 20t + 2$ metres after $t$ seconds. Find its average velocity from $t = 1$ to $t = 3$ and its instantaneous velocity at $t = 1$.

Average $0.4$ m/s; instantaneous $10.2$ m/s at $t = 1$

For $f(x) = \frac{1}{x + 1}$, find the exact instantaneous rate of change at $x = 1$ and the equation of the tangent line there.

Rate $-\frac{1}{4}$; tangent $y = -\frac{1}{4}x + \frac{3}{4}$

Grade 12 · MHF4U · Advanced Functions

MHF4U: Average vs Instantaneous Rate of Change

A rate of change says how fast one quantity changes compared with another: metres per second, dollars per item. An average rate covers an interval. An instantaneous rate is the rate at one moment, and MHF4U estimates it by taking the average over smaller and smaller intervals.

How It Works

Average rate of change from $x = a$ to $x = b$ is $\frac{f(b) - f(a)}{b - a}$ , the slope of the secant line through the two points on the graph. Its units are the units of $f$ per unit of $x$ . For a height in metres and a time in seconds, it is an average velocity in m/s.

Instantaneous rate of change at $x = a$ is the slope of the tangent line there. A tangent touches the graph at one point, so you cannot take its slope from two points on it. Instead, take a second point close to $a$ , at $a + h$ , and work out the secant slope $\frac{f(a + h) - f(a)}{h}$ . As $h$ shrinks, $0.1$ , then $0.01$ , then $0.001$ , the secant slopes settle on one value: that value is the instantaneous rate.

Doing it algebraically. Expand $f(a + h)$ , subtract $f(a)$ , and divide by $h$ . The $h$ in the denominator cancels, and what is left is a simple expression in $h$ . Let $h$ go to $0$ in that expression to read off the exact rate. With a table of values, use the two points closest to $a$ , one on each side, for the best estimate.

Average and instantaneous rates can be very different. A ball thrown up and caught at about the same height has an average velocity near $0$ over the flight, while its instantaneous velocity is large going up and large going down.

A positive rate means the function is increasing there, a negative rate means it is decreasing, and a smooth graph can only turn where the rate is $0$ .

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Find the average rate of change of $f(x) = x^2 - 3x$ from $x = 1$ to $x = 4$ .
Show solution
1. $f(4) = 16 - 12 = 4$ and $f(1) = 1 - 3 = -2$ .
2. $\frac{4 - (-2)}{4 - 1} = \frac{6}{3} = 2$ .
Answer: $2$
Easy
Find the average rate of change of $f(x) = 2^x$ on $0 \le x \le 3$ .
Show solution
1. $f(3) = 8$ and $f(0) = 1$ .
2. $\frac{8 - 1}{3 - 0} = \frac{7}{3}$ .
Answer: $\frac{7}{3} \approx 2.33$
Medium
Estimate the instantaneous rate of change of $f(x) = x^3$ at $x = 2$ using $h = 0.1$ , $0.01$ and $0.001$ .
Show solution
1. The secant slope is $\frac{(2 + h)^3 - 8}{h}$ .
2. $h = 0.1$ : $\frac{9.261 - 8}{0.1} = 12.61$ . $h = 0.01$ : $\frac{8.120601 - 8}{0.01} = 12.0601$ . $h = 0.001$ : $12.006001$ .
3. The slopes approach $12$ .
Answer: About $12$
Medium
A ball's height is $h(t) = -4.9t^2 + 20t + 2$ metres after $t$ seconds. Find its average velocity from $t = 1$ to $t = 3$ and its instantaneous velocity at $t = 1$ .
Show solution
1. $h(1) = 17.1$ and $h(3) = 17.9$ , so the average velocity is $\frac{17.9 - 17.1}{2} = 0.4$ m/s.
2. At $t = 1$ : $\frac{h(1 + h) - h(1)}{h}$ . With $h(1 + h) = -4.9(1 + 2h + h^2) + 20 + 20h + 2 = 17.1 + 10.2h - 4.9h^2$ , the quotient is $10.2 - 4.9h$ .
3. As $h \to 0$ this goes to $10.2$ .
Answer: Average $0.4$ m/s; instantaneous $10.2$ m/s at $t = 1$
Test level
For $f(x) = \frac{1}{x + 1}$ , find the exact instantaneous rate of change at $x = 1$ and the equation of the tangent line there.
Show solution
1. $f(1) = \frac{1}{2}$ and $f(1 + h) = \frac{1}{2 + h}$ .
2. $\frac{f(1 + h) - f(1)}{h} = \frac{1}{h}\left(\frac{1}{2 + h} - \frac{1}{2}\right) = \frac{1}{h} \cdot \frac{2 - (2 + h)}{2(2 + h)} = \frac{-1}{2(2 + h)}$ .
3. As $h \to 0$ this goes to $-\frac{1}{4}$ .
4. Tangent through $\left(1, \frac{1}{2}\right)$ with slope $-\frac{1}{4}$ : $y - \frac{1}{2} = -\frac{1}{4}(x - 1)$ , so $y = -\frac{1}{4}x + \frac{3}{4}$ .
Answer: Rate $-\frac{1}{4}$ ; tangent $y = -\frac{1}{4}x + \frac{3}{4}$

Learn It on Martian Lab

Martian Lab teaches this in the Rate of Change lesson of its MHF4U Polynomial Functions unit. Watch it, then take the Combining Functions practice tests to see if it stuck.

Watch the Rate of Change lesson free

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More Combining Functions topics: Sums, Products and Quotients of Functions · Composition of Functions

More MHF4U topics: Long and Synthetic Division · Vertical and Horizontal Asymptotes · Proving Trig Identities · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations · End Behaviour and Degree · Finding the Equation of a Polynomial · Even and Odd Functions · Remainder and Factor Theorem · Solving Polynomial Equations · Polynomial Inequalities · Sum and Difference of Cubes · Holes in Rational Functions · Graphing Rational Functions · Rational Equations and Inequalities · Radian Measure · Unit Circle and Special Angles · Compound Angle Formulas · Double Angle Formulas · Graphing Sine and Cosine with Transformations · Graphs of Tan, Sec, Csc and Cot · Sinusoidal Word Problems · Solving Exponential Equations · Graphing Logarithmic Functions · Exponential and Log Word Problems

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