Interpretation — instantaneous rate of change, slope of tangent: Exercises

Before we start, a picture of the one move that all L2 problems share — the secant sliding into the tangent.

Figure — Interpretation — instantaneous rate of change, slope of tangent

L1 — Recognition

Goal: recognise which object is which. No heavy algebra.

Q1. A car's average speed over a 2-hour trip was $60$ km/h. On the graph of distance vs. time, which object has slope $60$ : the tangent at $t = 1$ h, or the secant joining the start and end points? Explain in one line.

Recall Solution Q1

The secant joining start $(0, d_{0})$ to end $(2, d_{0} + 120)$ . Average speed $= \frac{total distance}{total time} = \frac{Δ d}{Δ t}$ , which is exactly rise over run between two real points — the definition of a secant slope (see Average rate of change). The tangent at $t = 1$ is the instantaneous speed, generally a different number.

Q2. For $f (x) = x^{2}$ , write the difference quotient at $x = a$ without simplifying it. Then say what quantity it represents geometrically for a fixed $h \neq = 0$ .

Recall Solution Q2

$\frac{f ( a + h ) - f ( a )}{h} = \frac{( a + h ) ^{2} - a ^{2}}{h}$ For a fixed $h \neq = 0$ this is the slope of the secant line through $(a, a^{2})$ and $(a + h, (a + h)^{2})$ — the average rate of change over the interval $[a, a + h]$ . It becomes the derivative only after the limit $h \to 0$ .

Q3. True or false: " $f^{'} (a)$ is the slope of the tangent line at $(a, f (a))$ and the rate at which $f$ changes per unit $x$ at that point." Justify.

Recall Solution Q3

True. Both are $\frac{change in output}{change in input}$ at the instant. The tangent is the best straight-line copy of the curve near $a$ ; its slope = rise/run = rate. Geometry and rate are one number in two costumes.

L2 — Application

Goal: run the four-step machine (build → expand → cancel $h$ → limit) end to end.

Q4. Using the limit definition (no shortcut rules), find $f^{'} (2)$ for $f (x) = x^{2} - 4 x + 1$ .

Recall Solution Q4

Step 1 (build): $\frac{f ( 2 + h ) - f ( 2 )}{h}$ . Here $f (2) = 4 - 8 + 1 = - 3$ . Step 2 (expand): $f (2 + h) = (2 + h)^{2} - 4 (2 + h) + 1 = 4 + 4 h + h^{2} - 8 - 4 h + 1 = h^{2} - 3$ . So numerator $= (h^{2} - 3) - (- 3) = h^{2}$ . Step 3 (cancel $h$ ): $\frac{h ^{2}}{h} = h$ for $h \neq = 0$ . We remove the illegal $0/0$ before limiting. Step 4 (limit): $h \to 0 lim h = 0$ . So $f^{'} (2) = 0$ . Meaning: the tangent at $x = 2$ is horizontal — $x = 2$ is the vertex of this parabola.

Q5. Find the equation of the tangent line to $g (x) = x^{2}$ at $x = 3$ , given from the parent note that $g^{'} (3) = 6$ .

Recall Solution Q5

Point: $(3, g (3)) = (3, 9)$ . Slope: $g^{'} (3) = 6$ . Point–slope form (a line needs a point + a slope): $y - 9 = 6 (x - 3) ⟹ y = 6 x - 9 .$

Q6. A ball's height is $s (t) = 5 t^{2}$ metres. Using the limit definition, find the instantaneous velocity at $t = 3$ s.

Recall Solution Q6

$s (3) = 45$ . $s (3 + h) = 5 (3 + h)^{2} = 5 (9 + 6 h + h^{2}) = 45 + 30 h + 5 h^{2}$ . $\frac{s ( 3 + h ) - s ( 3 )}{h} = \frac{30 h + 5 h ^{2}}{h} = 30 + 5 h (h \neq = 0) .$ $h \to 0 lim (30 + 5 h) = 30$ . So $v (3) = 30 m/s$ (see Velocity and acceleration).

Q7. Find $f^{'} (a)$ for $f (x) = \frac{1}{x}$ at a general $a \neq = 0$ from the definition.

Recall Solution Q7

Step 1: $\frac{\frac{1}{a + h} - \frac{1}{a}}{h}$ . Step 2 (common denominator): $\frac{1}{a + h} - \frac{1}{a} = \frac{a - ( a + h )}{a ( a + h )} = \frac{- h}{a ( a + h )}$ . Step 3 (divide by $h$ ): $\frac{- h}{a ( a + h )} \cdot \frac{1}{h} = \frac{- 1}{a ( a + h )}$ for $h \neq = 0$ . Step 4 (limit): $h \to 0 lim \frac{- 1}{a ( a + h )} = \frac{- 1}{a ^{2}}$ . So $f^{'} (a) = - \frac{1}{a ^{2}}$ — always negative, so $1/ x$ always falls (for $a \neq = 0$ ).

L3 — Analysis

Goal: reason about existence, sign, and behaviour — not just compute.

Q8. For $f (x) = ∣ x - 1∣$ , decide whether $f^{'} (1)$ exists. Show the left and right computations.

Recall Solution Q8

Difference quotient at $a = 1$ : $\frac{∣ ( 1 + h ) - 1∣ - 0}{h} = \frac{∣ h ∣}{h}$ .

From the right ( $h > 0$ ): $∣ h ∣ = h$ , ratio $= + 1$ .
From the left ( $h < 0$ ): $∣ h ∣ = - h$ , ratio $= - 1$ .

Right limit $+ 1 \neq = - 1$ left limit, so the two-sided limit does not exist $\Rightarrow$ $f^{'} (1)$ does not exist. Geometrically there is a corner at $(1, 0)$ : no single tangent slope (see Differentiability and continuity). Look at the corner figure below.

Q9. The parabola $f (x) = x^{2}$ has secant slope $2 a + h$ over $[a, a + h]$ (you may take this as given). Explain, using this expression, why the tangent slope at $a$ must be $2 a$ — and what role the term $h$ plays.

Recall Solution Q9

Secant slope $= 2 a + h$ is built from two costs: a true part $2 a$ (fixed, doesn't depend on the interval) and an error part $h$ (the width of the interval). As we slide the second point in, $h \to 0$ , the error vanishes and only the true part survives: $f^{'} (a) = lim_{h \to 0} (2 a + h) = 2 a .$ So the tangent slope is exactly the "interval-independent skeleton" of the secant slope. This is the Power Rule result $\frac{d}{d x} x^{2} = 2 x$ , seen from first principles.

Q10. For $f (x) = x^{3}$ , its tangent at the origin is the line $y = 0$ . Is it true that "a tangent line touches the curve at exactly one point"? Investigate at the origin.

Recall Solution Q10

Slope at $0$ : difference quotient $\frac{( 0 + h ) ^{3} - 0}{h} = \frac{h ^{3}}{h} = h^{2} \to 0$ , so $f^{'} (0) = 0$ and the tangent is $y = 0$ (the $x$ -axis). But $y = 0$ meets $y = x^{3}$ wherever $x^{3} = 0$ — only at the origin here, yet the tangent crosses through the curve rather than resting outside it. In general "touches once" is false: a tangent is defined by matching slope locally (the derivative), not by counting intersections. (A tangent to $y = x^{3} - x$ at one point can hit the curve elsewhere.)

L4 — Synthesis

Goal: combine slope, tangent line, and rate ideas in one problem.

Q11. For $f (x) = x^{2} - 4 x + 1$ , find every point where the tangent line is horizontal, and confirm your answer matches Q4.

Recall Solution Q11

First get the general slope. Difference quotient: $\frac{( x + h ) ^{2} - 4 ( x + h ) + 1 - ( x ^{2} - 4 x + 1 )}{h} = \frac{2 x h + h ^{2} - 4 h}{h} = 2 x - 4 + h .$ Limit: $f^{'} (x) = 2 x - 4$ . Horizontal tangent means slope $0$ : $2 x - 4 = 0 \Rightarrow x = 2$ . Point: $f (2) = - 3$ , so the horizontal tangent is at $(2, - 3)$ , tangent line $y = - 3$ . This matches Q4 ( $f^{'} (2) = 0$ ).

Q12. A ball thrown up has height $s (t) = 20 t - 5 t^{2}$ metres. (a) Find its velocity function $v (t) = s^{'} (t)$ from the definition. (b) At what time is the ball momentarily at rest (top of flight)? (c) What is that maximum height?

Recall Solution Q12

(a) $s (t + h) = 20 (t + h) - 5 (t + h)^{2} = 20 t + 20 h - 5 t^{2} - 10 t h - 5 h^{2}$ . $\frac{s ( t + h ) - s ( t )}{h} = \frac{20 h - 10 t h - 5 h ^{2}}{h} = 20 - 10 t - 5 h .$ Limit: $v (t) = s^{'} (t) = 20 - 10 t$ m/s. (b) At rest $\Rightarrow v = 0$ : $20 - 10 t = 0 \Rightarrow t = 2$ s. (c) $s (2) = 40 - 20 = 20$ m. So $v (t) = 20 - 10 t, t = 2 s, h_{m a x} = 20 m$ . Note $v > 0$ (rising) for $t < 2$ , $v < 0$ (falling) for $t > 2$ — the sign of the derivative reports direction of motion (see Velocity and acceleration).

L5 — Mastery

Goal: prove a general statement and design a case.

Q13. From the limit definition, prove that for any constants $m, c$ the line $f (x) = m x + c$ has $f^{'} (a) = m$ for every $a$ . Interpret: what does this say about "average vs. instantaneous" for lines?

Recall Solution Q13

$\frac{f ( a + h ) - f ( a )}{h} = \frac{[ m ( a + h ) + c ] - [ ma + c ]}{h} = \frac{mh}{h} = m (h \neq = 0) .$ The quotient equals $m$ before any limit — it has no $h$ left. So $h \to 0 lim m = m$ , giving $f^{'} (a) = m$ for all $a$ . Interpretation: for a straight line the secant slope is already $m$ regardless of the interval, so average rate = instantaneous rate = $m$ everywhere. This is precisely why the L1 trap (confusing the two) only ever works for lines.

Q14. Design a function that is continuous at $x = 0$ but not differentiable there, different from $∣ x ∣$ , and prove it fails the derivative test. (Hint: a "vertical tangent" also kills the two-sided derivative.)

Recall Solution Q14

Take $f (x) = x^{1/3}$ (the cube root). It is continuous everywhere (no jump). Difference quotient at $0$ : $\frac{f ( 0 + h ) - f ( 0 )}{h} = \frac{h ^{1/3} - 0}{h} = \frac{h ^{1/3}}{h} = h^{- 2/3} = \frac{1}{h ^{2/3}} .$ As $h \to 0$ , $h^{2/3} \to 0^{+}$ (it's a square-type power, always $\geq 0$ ), so $\frac{1}{h ^{2/3}} \to + \infty$ . The limit is infinite from both sides, so $f^{'} (0)$ does not exist — the tangent is vertical (infinite slope). This is a second, distinct way differentiability can fail (corner in Q8 vs. vertical tangent here). Both keep continuity; both break the finite-slope limit (see Differentiability and continuity).

Recall One-glance answer key

Q4 $f^{'} (2) = 0$ · Q5 $y = 6 x - 9$ · Q6 $v (3) = 30$ m/s · Q7 $f^{'} (a) = - 1/ a^{2}$ · Q8 DNE (corner) · Q9 slope $= 2 a$ · Q10 "touches once" false · Q11 $(2, - 3)$ · Q12 $v (t) = 20 - 10 t$ , $t = 2$ s, $h_{m a x} = 20$ m · Q13 $f^{'} (a) = m$ · Q14 $f^{'} (0)$ DNE (vertical tangent).

Connections

Average rate of change — the secant slopes we limit in every L2 problem.
Limits — formal definition — the " $h \to 0$ " engine used throughout.
Derivative as a function — Q11/Q12 find $f^{'} (x)$ for all $x$ , not one point.
Power Rule — the shortcut Q9 proves from scratch for $x^{2}$ .
Differentiability and continuity — Q8, Q14: corner and vertical-tangent failures.
Velocity and acceleration — Q6, Q12: rate-of-change physics.