Lesson 1.7 · Unit 1 · F-IF.6
The slope between two points, and what happens as they get closer
The average rate of change is the slope of the line through two points of a graph. Shrink the gap between the points and that slope settles toward a single number, the instantaneous rate. That idea is the whole of calculus, and this lesson builds the algebra it needs.
- The average rate of change of \( f \) from \( a \) to \( b \) is \( \dfrac{f(b) - f(a)}{b - a} \).
- It is the slope of the secant line through \( \big(a, f(a)\big) \) and \( \big(b, f(b)\big) \).
- Its units are output units per input unit, such as meters per second or dollars per item.
- For a line it is the same everywhere; for a curve it depends on the interval.
- The difference quotient is \( \dfrac{f(x + h) - f(x)}{h} \), the average rate from \( x \) to \( x + h \).
- To simplify it, expand \( f(x + h) \), subtract \( f(x) \), and factor an \( h \) out of the numerator so it cancels.
- The \( h \) must cancel completely before you let \( h \) become small. If it cannot, there is an algebra error.
- A zero average rate does not mean no movement; it means the start and end values match.
Where students lose marks: writing \( f(x + h) \) as \( f(x) + h \) or \( f(x) + f(h) \). Substitute \( x + h \) for every \( x \) and expand. For \( f(x) = x^2 \), \( f(x + h) = x^2 + 2xh + h^2 \), which differs from both.
The problem. (a) Find the average rate of change of \( f(x) = x^2 - 3x \) on \( [1, 4] \). (b) Find it on \( [2, 2 + h] \) and see what happens as \( h \) shrinks. (c) Simplify the difference quotient of \( f \). (d) A ball's height is \( s(t) = -16t^2 + 64t \) feet. Find its average velocity on \( [1, 3] \) and interpret the result.
Step one: evaluate the endpoints for (a). \( f(1) = 1 - 3 = -2 \) and \( f(4) = 16 - 12 = 4 \).
Step two: form the quotient. \( \dfrac{4 - (-2)}{4 - 1} = \dfrac{6}{3} = 2 \). The secant line has slope 2.
Step three: expand for (b). \( f(2 + h) = (2 + h)^2 - 3(2 + h) = 4 + 4h + h^2 - 6 - 3h = h^2 + h - 2 \). And \( f(2) = 4 - 6 = -2 \).
Step four: form and simplify. \( \dfrac{f(2 + h) - f(2)}{h} = \dfrac{h^2 + h - 2 + 2}{h} = \dfrac{h^2 + h}{h} = h + 1 \). The \( h \) cancels completely.
Step five: let \( h \) shrink. For \( h = 1 \): 2. For \( h = 0.1 \): 1.1. For \( h = 0.01 \): 1.01. The values approach 1. So the slope at \( x = 2 \) is 1. Check with the vertex: the parabola \( x^2 - 3x \) has its vertex at \( x = 1.5 \), so at \( x = 2 \) it is rising slowly, consistent with a small positive slope.
Step six: simplify the general quotient for (c). \( f(x + h) = (x + h)^2 - 3(x + h) = x^2 + 2xh + h^2 - 3x - 3h \). Subtract \( f(x) = x^2 - 3x \): the \( x^2 \) and the \( -3x \) cancel, leaving \( 2xh + h^2 - 3h \). Divide by \( h \): \( 2x + h - 3 \). Check with \( x = 2 \): \( 4 + h - 3 = h + 1 \), matching step four.
Step seven: evaluate (d). \( s(1) = -16 + 64 = 48 \) and \( s(3) = -144 + 192 = 48 \). Average velocity: \( \dfrac{48 - 48}{3 - 1} = 0 \) feet per second.
Step eight: interpret it. The ball is moving the whole time, rising to its peak at \( t = 2 \) and falling back, but it is at the same height at \( t = 1 \) and \( t = 3 \). An average over an interval hides what happens inside it. Shrinking the interval reveals the truth: on \( [1, 1.01] \), \( s(1.01) = 48.3184 \), so the average velocity is \( \dfrac{0.3184}{0.01} = 31.84 \) ft/s, close to 32. The instantaneous velocity at \( t = 1 \) is 32 ft/s upward, and the calculus of Unit 11 turns that shrinking into a definition.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- Find the average rate of change of \( f(x) = 3x + 1 \) on \( [1, 5] \).
Show the full solution
\( f(1) = 4 \) and \( f(5) = 16 \), so \( \dfrac{12}{4} = 3 \). A line has the same rate everywhere. 3
- Find the average rate of change of \( f(x) = x^2 \) on \( [1, 3] \).
Show the full solution
\( \dfrac{9 - 1}{3 - 1} = \dfrac{8}{2} \). 4
- Find the average rate of change of \( f(x) = x^2 - x \) on \( [0, 4] \).
Show the full solution
\( f(0) = 0 \) and \( f(4) = 12 \), so \( \dfrac{12}{4} \). 3
- Simplify the difference quotient for \( f(x) = 2x + 5 \).
Show the full solution
\( f(x + h) - f(x) = 2(x + h) + 5 - 2x - 5 = 2h \). Divide by \( h \). 2
- Simplify the difference quotient for \( f(x) = x^2 \).
Show the full solution
\( \dfrac{(x + h)^2 - x^2}{h} = \dfrac{2xh + h^2}{h} \). \( 2x + h \)
- Find the average rate of change of \( f(x) = \dfrac{1}{x} \) on \( [1, 4] \).
Show the full solution
\( f(1) = 1 \) and \( f(4) = \dfrac{1}{4} \), so \( \dfrac{\frac14 - 1}{3} = \dfrac{-\frac34}{3} = -\dfrac{1}{4} \). Negative because the function is decreasing. \( -\dfrac{1}{4} \)
- Simplify the difference quotient for \( f(x) = x^3 \).
Show the full solution
\( (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \). Subtract \( x^3 \) and divide by \( h \). Check with \( x = 1, h = 1 \): \( \dfrac{8 - 1}{1} = 7 \), and \( 3 + 3 + 1 = 7 \). Correct. \( 3x^2 + 3xh + h^2 \)
- For \( f(x) = \sqrt{x} \), find the average rate on \( [4, 9] \), then simplify
\( \dfrac{\sqrt{4 + h} - 2}{h} \) by rationalizing.
Show the full solution
On \( [4, 9] \): \( \dfrac{3 - 2}{5} = \dfrac{1}{5} \). For the quotient multiply by the conjugate: \( \dfrac{(\sqrt{4 + h} - 2)(\sqrt{4 + h} + 2)}{h(\sqrt{4 + h} + 2)} = \dfrac{(4 + h) - 4}{h(\sqrt{4 + h} + 2)} = \dfrac{1}{\sqrt{4 + h} + 2} \). As \( h \) becomes small this approaches \( \dfrac{1}{4} \). Check at \( h = 5 \): the first gives \( \dfrac{\sqrt{9} - 2}{5} = \dfrac{1}{5} \), and \( \dfrac{1}{3 + 2} = \dfrac15 \). \( \dfrac{1}{5} \); the quotient is \( \dfrac{1}{\sqrt{4 + h} + 2} \)
- Simplify the difference quotient for \( f(x) = \dfrac{1}{x} \).
Show the full solution
\( \dfrac{\frac{1}{x + h} - \frac{1}{x}}{h} \). Common denominator in the numerator: \( \dfrac{x - (x + h)}{x(x + h)} = \dfrac{-h}{x(x + h)} \). Divide by \( h \): \( -\dfrac{1}{x(x + h)} \). Check with \( x = 1, h = 1 \): \( \dfrac{\frac12 - 1}{1} = -\dfrac12 \), and \( -\dfrac{1}{1 \cdot 2} = -\dfrac12 \). Correct. \( -\dfrac{1}{x(x + h)} \)
- For \( s(t) = -16t^2 + 64t \), the average velocity on \( [1, 3] \) is 0. Explain what this
means, and find the average velocity on \( [1, 1.01] \) and the apparent instantaneous
velocity at \( t = 1 \).
Show the full solution
An average of zero means the ball ends at the height where it began over that interval. It does not mean the ball was at rest: it rose and fell. \( s(1) = 48 \) and \( s(1.01) = -16(1.0201) + 64.64 = 48.3184 \). Average velocity: \( \dfrac{0.3184}{0.01} = 31.84 \) ft/s. The exact difference quotient is \( -32t - 16h + 64 \), which at \( t = 1 \) is \( 32 - 16h \); for \( h = 0.01 \) that is 31.84, matching, and as \( h \to 0 \) it approaches 32. Zero average, but 31.84 ft/s on the short interval and 32 ft/s at the instant