Unit 10, lesson 111

Difference Quotients and the Instantaneous-Rate Idea

How to write and simplify the difference quotient with h kept nonzero, how shrinking secant lines lead to a rate at a single instant when that limit exists, and why a corner has no single tangent slope.

  • Theory part 221
  • Practice part 222
  • Rates and Areas

Water drains from a tank. You read the level at noon and again ten minutes later, and the drop divided by ten minutes is an average rate. But how fast is the level falling at noon exactly? Shrink the ten minutes to one minute, then to one second, and watch where the averages go. Done carefully with algebra, that is the idea behind the derivative in calculus.

Key terms

Difference quotient

Definition Let ff be a function, let aa be in its domain, and let h≠0h \ne 0 be a number with a+ha + h also in the domain. The difference quotient of ff at aa is

f(a+h)−f(a)h.\frac{f(a + h) - f(a)}{h}.

It is the average rate of change of ff from aa to a+ha + h, and the slope of the secant line through (a,f(a))(a, f(a)) and (a+h,f(a+h))(a + h, f(a + h)). With a variable xx in place of aa it is written

f(x+h)−f(x)h.\frac{f(x + h) - f(x)}{h}.

In plain words Step from aa to a+ha + h, see how much the output changed, and divide by the size of the step.

Example Take f(x)=3x2−xf(x) = 3x^2 - x and a=2a = 2, so f(2)=10f(2) = 10. With h=0.5h = 0.5 the second point is f(2.5)=18.75−2.5=16.25f(2.5) = 18.75 - 2.5 = 16.25, and

f(2.5)−f(2)0.5=16.25−100.5=12.5.\begin{aligned} \frac{f(2.5) - f(2)}{0.5} &= \frac{16.25 - 10}{0.5} \\[0.8em] &= 12.5. \end{aligned}

Nonzero h

Definition The step hh in a difference quotient must not be 00. At h=0h = 0 the quotient becomes f(a)−f(a)0=00\frac{f(a) - f(a)}{0} = \frac{0}{0}, which is not a number. The step may be negative, and then a+ha + h lies to the left of aa. Any simplified form of the quotient equals it only for h≠0h \ne 0, so this condition has to be carried along through every step.

In plain words You cannot measure a rate over a step of length zero. Small steps are fine, and steps to the left are fine, but the step itself is never zero.

Example For f(x)=3x2−xf(x) = 3x^2 - x at a=2a = 2, expanding gives f(2+h)=10+11h+3h2f(2 + h) = 10 + 11h + 3h^2. For h≠0h \ne 0

f(2+h)−f(2)h=11h+3h2h=11+3h.\begin{aligned} \frac{f(2 + h) - f(2)}{h} &= \frac{11h + 3h^2}{h} \\[0.8em] &= 11 + 3h. \end{aligned}

The expression 11+3h11 + 3h has a value at h=0h = 0, but the quotient does not. With the negative step h=−0.5h = -0.5 the secant runs through (1.5,5.25)(1.5, 5.25) and (2,10)(2, 10), and its slope is 5.25−10−0.5=9.5\frac{5.25 - 10}{-0.5} = 9.5, which matches 11+3(−0.5)=9.511 + 3(-0.5) = 9.5.

Original domain

Definition The difference quotient of ff at aa is defined exactly for the numbers hh with h≠0h \ne 0 and a+ha + h in the domain of ff. These restrictions come from the original function and stay in force after any simplification, even when the simplified formula has no visible trouble at those values of hh.

In plain words Both points of the secant have to be on the graph. Before simplifying, write down which steps hh are allowed, and keep that list.

Example For f(x)=xf(x) = \sqrt{x} at a=4a = 4 the second point needs 4+h≥04 + h \ge 0, so the allowed steps are h≥−4h \ge -4 with h≠0h \ne 0. The step h=−5h = -5 is not allowed, since it asks for −1\sqrt{-1}, which is not a real number. For f(x)=1xf(x) = \frac{1}{x} at a=2a = 2 the second point needs 2+h≠02 + h \ne 0, so the allowed steps are all hh except 00 and −2-2.

Algebraic simplification

Definition To see what a difference quotient does for small hh, rewrite it, for the allowed hh, in a form without the factor hh in the denominator. The usual moves are expanding a polynomial and cancelling hh, combining fractions over a common denominator, and multiplying the numerator and the denominator by a conjugate when there is a square root. Cancelling hh is valid because h≠0h \ne 0, and the simplified form equals the quotient only for the allowed hh.

In plain words The goal is to get rid of the hh in the denominator, since that is what makes h=0h = 0 impossible to plug in. Algebra does this without changing any values where the quotient is defined.

Example For f(x)=xf(x) = \sqrt{x} at a=4a = 4, multiply by the conjugate 4+h+2\sqrt{4 + h} + 2, which is positive and so never zero. For h≥−4h \ge -4 with h≠0h \ne 0,

4+h−2h=(4+h)−4h(4+h+2)=14+h+2.\begin{aligned} \frac{\sqrt{4 + h} - 2}{h} &= \frac{(4 + h) - 4}{h\left(\sqrt{4 + h} + 2\right)} \\[0.8em] &= \frac{1}{\sqrt{4 + h} + 2}. \end{aligned}

The last form is defined at h=0h = 0, where it equals 14\frac{1}{4}, but the quotient is still undefined there. The restrictions h≥−4h \ge -4 and h≠0h \ne 0 stay.

Shrinking secants

Definition As hh takes values closer and closer to 00, from both sides and never equal to 00, the secant line through (a,f(a))(a, f(a)) and (a+h,f(a+h))(a + h, f(a + h)) pivots about the fixed point (a,f(a))(a, f(a)). When the graph has no break at aa, the second point slides along the graph toward (a,f(a))(a, f(a)). The secant slopes are the difference quotients, so their behavior is the behavior of the quotient as hh approaches 00.

In plain words Pin one end of a line to the graph at aa and slide the other end toward it along the curve. Watch what happens to the slope.

Example For f(x)=3x2−xf(x) = 3x^2 - x at a=2a = 2 the secant slopes are 11+3h11 + 3h.

hh110.10.10.010.01−0.01-0.01−0.1-0.1−1-1
Secant slope141411.311.311.0311.0310.9710.9710.710.788

From both sides the slopes close in on 1111. The line through (2,10)(2, 10) with slope 1111 is y=10+11(x−2)y = 10 + 11(x - 2), or y=11x−12y = 11x - 12, the line the secants settle into.

Instantaneous rate as a limit, when it exists

Definition Let ff be defined on an open interval containing aa. If the difference quotients approach a single real number mm as hh approaches 00 from both sides, that is

m=lim⁡h→0f(a+h)−f(a)h,m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h},

then mm is the instantaneous rate of change of ff at aa, and the line through (a,f(a))(a, f(a)) with slope mm is the tangent line there. If the quotients do not approach a single real number, ff has no instantaneous rate of change at aa. Calculus calls mm the derivative of ff at aa.

In plain words The rate at one instant is the number the average rates close in on as the time step shrinks. It exists only if they really do close in on one number.

Example A ball thrown upward has height s(t)=20t−5t2s(t) = 20t - 5t^2 meters after tt seconds, so s(1)=15s(1) = 15. For h≠0h \ne 0

s(1+h)−s(1)h=10h−5h2h=10−5h.\begin{aligned} \frac{s(1 + h) - s(1)}{h} &= \frac{10h - 5h^2}{h} \\[0.8em] &= 10 - 5h. \end{aligned}

As hh approaches 00 this approaches 1010, so at t=1t = 1 the ball is rising at 1010 meters per second.

Corners and the failure of a single tangent slope

Definition Suppose the difference quotients of ff at aa approach one number m1m_1 as hh approaches 00 through negative values and a different number m2m_2 as hh approaches 00 through positive values. Then the two-sided limit does not exist, and the graph has a corner at (a,f(a))(a, f(a)). The graph has no break there, since f(a+h)−f(a)f(a + h) - f(a) is hh times the quotient and so approaches 00. Yet it has no single tangent slope and ff has no instantaneous rate of change at aa. A single tangent slope can also fail at a jump or where the quotients grow without bound.

In plain words At a corner the secants from the left settle on one slope and the secants from the right settle on another. No single line fits both, so the curve has no one slope at that point.

Example Take f(x)=∣x2−4∣f(x) = \lvert x^2 - 4 \rvert at a=2a = 2, where f(2)=0f(2) = 0. For h>0h > 0 the number (2+h)2−4=h(4+h)(2 + h)^2 - 4 = h(4 + h) is positive, and for −4<h<0-4 < h < 0 it is negative, so

f(2+h)−f(2)h={4+h,h>0,−(4+h),−4<h<0.\begin{aligned} &\frac{f(2 + h) - f(2)}{h} \\[0.8em] &\quad = \begin{cases} 4 + h, & h > 0, \\ -(4 + h), & -4 < h < 0. \end{cases} \end{aligned}
hh0.10.10.010.01−0.01-0.01−0.1-0.1
Secant slope4.14.14.014.01−3.99-3.99−3.9-3.9

From the right the slopes approach 44 and from the left they approach −4-4. The graph is unbroken at (2,0)(2, 0) but has a sharp corner there, and ff has no instantaneous rate of change at 22. Compare the smooth curve 3x2−x3x^2 - x above, whose slopes approach 1111 from both sides.

Check yourself

  1. Simplify the difference quotient of f(x)=x2−5xf(x) = x^2 - 5x at a=3a = 3 and find the instantaneous rate of change there.
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Here f(3)=−6f(3) = -6 and f(3+h)=−6+h+h2f(3 + h) = -6 + h + h^2. For h≠0h \ne 0

f(3+h)−f(3)h=h+h2h=1+h,\begin{aligned} \frac{f(3 + h) - f(3)}{h} &= \frac{h + h^2}{h} \\[0.8em] &= 1 + h, \end{aligned}

which approaches 11. The instantaneous rate of change at 33 is 11.

  1. For f(x)=1x+1f(x) = \frac{1}{x + 1} at a=1a = 1, which steps hh are allowed? Simplify the difference quotient and find its limit as hh approaches 00.
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The second point needs 2+h≠02 + h \ne 0, so every hh except 00 and −2-2 is allowed. Since f(1)=12f(1) = \frac{1}{2}, for those hh

12+h−12h=2−(2+h)2h(2+h)=−12(2+h),\begin{aligned} \frac{\frac{1}{2 + h} - \frac{1}{2}}{h} &= \frac{2 - (2 + h)}{2h(2 + h)} \\[0.8em] &= -\frac{1}{2(2 + h)}, \end{aligned}

which approaches −14-\frac{1}{4}.

  1. Does f(x)=∣2x−6∣f(x) = \lvert 2x - 6 \rvert have an instantaneous rate of change at x=3x = 3?
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No. Since f(3)=0f(3) = 0 and f(3+h)=∣2h∣f(3 + h) = \lvert 2h \rvert, the quotient is ∣2h∣h\frac{\lvert 2h \rvert}{h}, which equals 22 for every h>0h > 0 and −2-2 for every h<0h < 0. The one-sided slopes differ, so the graph has a corner at (3,0)(3, 0).

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