The average rate of change of f on [a, b] is the slope of the secant line:
average rate of change = [f(b) − f(a)] / (b − a)
For f(x) = x² + 1 on [1, 4]: [f(4) − f(1)]/(4 − 1) = (17 − 2)/3 = 5.
The instantaneous rate of change at a single point x = a is the slope of the tangent line there. It answers "how fast is f changing at the exact instant x = a?" — the quantity the derivative was invented to capture.
Step a tiny distance h away from x. The secant slope through (x, f(x)) and (x + h, f(x + h)) is the difference quotient. Its limit as h → 0 is the derivative:
f'(x) = lim_{h→0} [f(x + h) − f(x)] / h
When this limit exists, f is differentiable at x, and f'(x) is the slope of the tangent line at (x, f(x)). Worked once, for f(x) = x²:
[(x + h)² − x²] / h = (2xh + h²)/h = 2x + h → lim_{h→0} (2x + h) = 2x
There is an equivalent at-a-point form, handy when a fixed point a is given:
f'(a) = lim_{x→a} [f(x) − f(a)] / (x − a)
CLEP loves a limit that is a derivative, dressed up so you must recognize the pattern. Match the expression to lim_{x→a} [f(x) − f(a)]/(x − a) = f'(a) or lim_{h→0}[f(a+h) − f(a)]/h = f'(a).
lim_{h→0} [(2 + h)³ − 8] / h = d/dx[x³] at x = 2 = 3(2)² = 12
lim_{x→π} (ln x − ln π)/(x − π) = d/dx[ln x] at x = π = 1/π
Reading these as derivatives is far faster than trying to manipulate the fraction directly.
The derivative gives the slope; the point comes from the function. In point-slope form, the tangent line at x = a is
y − f(a) = f'(a)·(x − a) equivalently L(x) = f(a) + f'(a)(x − a)
For f(x) = x² − 4x at x = 1: f(1) = −3, f'(x) = 2x − 4, f'(1) = −2, so y = −2x − 1.
Because a curve hugs its tangent line near the point of tangency, L(x) gives a fast hand approximation of f(x) — linearization. To estimate √25.3, use f(x) = √x at a = 25: f(25) = 5, f'(x) = 1/(2√x), f'(25) = 1/10, so
L(25.3) = 5 + (1/10)(0.3) = 5.03 (true value ≈ 5.0299)
Over- or underestimate? The sign of the concavity decides which side of the curve the tangent line sits on:
- Concave down (f'' < 0): the curve bends below its tangent, so L overestimates.
- Concave up (f'' > 0): the curve bends above its tangent, so L underestimates.
Since √x is concave down, 5.03 is an overestimate — consistent with 5.0299.
[GRAPH: f(x) = √x with tangent line L(x) = 5 + (1/10)(x−25) at x = 25, window [20, 30] × [4, 6]
- Curve concave down, passing through (25, 5)
- Straight tangent line touching at (25, 5), lying ABOVE the curve
- Point on line at (25.3, 5.03) sits just above point on curve at (25.3, 5.0299)
- Label: "concave down ⇒ tangent above curve ⇒ overestimate"]
With only data, approximate f'(a) using the closest values on each side (a symmetric difference):
f'(a) ≈ [f(a + h) − f(a − h)] / (2h)
x |
1 | 2 | 4 | 6 | 7 |
|---|---|---|---|---|---|
g(x) |
3 | 5 | 12 | 17 | 19 |
g'(4) ≈ [g(6) − g(2)]/(6 − 2) = (17 − 5)/4 = 3. This is always an estimate, never exact.
Theorem. If
fis differentiable atx = a, thenfis continuous atx = a.
The converse is false. A function can be continuous yet not differentiable. A derivative fails to exist at:
- a corner — the one-sided slopes are finite but unequal, e.g. |x − 3| at x = 3 (left slope −1, right slope +1);
- a cusp — the one-sided slopes run to +∞ and −∞, e.g. x^(2/3) at 0;
- a vertical tangent — the slope runs to ±∞ from both sides, e.g. x^(1/3) at 0;
- any discontinuity — no derivative where the graph breaks.
f'(a).f'(x) = lim_{h→0} [f(x + h) − f(x)]/h; recognize this pattern to read a "disguised" limit as a derivative.L(x) = f(a) + f'(a)(x − a); use it to approximate nearby values.lim_{h→0} [(2 + h)³ − 8] / h is1. E. Correct: this is d/dx[x³] at x = 2, i.e. 3(2)² = 12. A) 0 comes from wrongly plugging h = 0 into 0/0. B) 3 uses the exponent only. C) 6 is 2·3, a miscount. D) 8 is f(2), not the slope. Fix rule: [(a+h)ⁿ − aⁿ]/h → f'(a).
lim_{x→π} (ln x − ln π)/(x − π) is2. E. Correct: this matches [f(x) − f(a)]/(x − a) with f = ln x, a = π, giving f'(π) = 1/π. A) 0 misreads it as 0/0. B) 1 forgets to evaluate 1/x at π. C) ln π is a value, not a slope. D) π inverts the derivative. Fix rule: (f(x) − f(a))/(x − a) → f'(a).
f(x) = x² + 1 on [1, 4] is3. A. Correct: (f(4) − f(1))/(4 − 1) = (17 − 2)/3 = 5. B) 3 is the interval width. C) 15 is the numerator only. D) 17 is f(4). E) 8 mis-subtracts. Fix rule: average rate = output change over input change.
f(x) = x² − 4x at x = 1 is4. C. Correct: f(1) = −3, f'(1) = 2(1) − 4 = −2, so y + 3 = −2(x − 1) gives y = −2x − 1. A) flips the constant sign. B) uses the wrong slope. D) mis-distributes. E) uses f values incorrectly. Fix rule: point from f(a), slope from f'(a).
f(x) = √x at x = 25, the approximation of √25.3 is5. A. Correct: L(25.3) = 5 + (1/10)(0.3) = 5.03. B) 5.3 misplaces the decimal. C) 5.0299 is the true value, not the linear estimate. D) 5.15 uses slope 1/2. E) 5.006 mis-scales 0.3. Fix rule: L(x) = f(a) + f'(a)(x − a).
From the table, the best estimate of g'(4) is
x |
1 | 2 | 4 | 6 | 7 |
|---|---|---|---|---|---|
g(x) |
3 | 5 | 12 | 17 | 19 |
6. C. Correct: symmetric difference [g(6) − g(2)]/(6 − 2) = (17 − 5)/4 = 3. A) 6 uses a one-sided [g(6) − g(4)]/(6 − 4). B) 5 mis-subtracts. D) 2 divides by the wrong width. E) 3.5 averages the wrong pairs. Fix rule: center the interval on the point.
f is differentiable at x = a, which must be true?7. D. Correct: differentiable ⇒ continuous. A) a differentiable function need not pass through 0. B) and C) a nonzero slope is fine. E) differentiability at a does not force a second derivative. Fix rule: differentiability is the stronger condition and guarantees continuity.
√25.3 ≈ 5.03 is an8. D. Correct: √x has f''(x) = −1/(4x^{3/2}) < 0 (concave down), so the tangent line lies above the curve and 5.03 overestimates. A) and C) wrongly call the estimate low. B) the estimate is not exact. E) √x is not concave up. Fix rule: concave down → tangent above → overestimate.
x = 3, the function f(x) = |x − 3| is9. B. Correct: |x − 3| is continuous at 3 (both pieces equal 0) but has a corner — left slope −1, right slope +1 — so it is not differentiable. A) and D) no single slope exists. C) the graph is unbroken. E) it is continuous. Fix rule: a corner is continuous but not differentiable.
f(x) = x², lim_{h→0} [(x + h)² − x²]/h equals10. D. Correct: [(x+h)² − x²]/h = 2x + h → 2x. A) 0 plugs in h = 0 prematurely. B) x halves the coefficient. C) x² is the function, not its rate. E) 2x + h forgets the limit. Fix rule: simplify the difference quotient, then let h → 0.
lim_{h→0} [f(a + h) − f(a)]/h represents11. B. Correct: this is the limit definition of the derivative at a. A) and D) describe the difference quotient (a secant slope) before the limit. C) f(a) is a value. E) the limit exists whenever f is differentiable at a. Fix rule: the limit of the difference quotient is f'(a).
f(x) = ln x at x = 1, the estimate of ln(1.1) is12. C. Correct: at a = 1, L(x) = ln 1 + 1·(x − 1) = x − 1, so L(1.1) = 0.1. A) 1.1 is the input. B) 0 is ln 1. D) 0.0953 is the true value, not the linear estimate. E) 1 is the slope. Fix rule: evaluate the tangent line, not the curve.
1. E. Correct: this is d/dx[x³] at x = 2, i.e. 3(2)² = 12. A) 0 comes from wrongly plugging h = 0 into 0/0. B) 3 uses the exponent only. C) 6 is 2·3, a miscount. D) 8 is f(2), not the slope. Fix rule: [(a+h)ⁿ − aⁿ]/h → f'(a).
2. E. Correct: this matches [f(x) − f(a)]/(x − a) with f = ln x, a = π, giving f'(π) = 1/π. A) 0 misreads it as 0/0. B) 1 forgets to evaluate 1/x at π. C) ln π is a value, not a slope. D) π inverts the derivative. Fix rule: (f(x) − f(a))/(x − a) → f'(a).
3. A. Correct: (f(4) − f(1))/(4 − 1) = (17 − 2)/3 = 5. B) 3 is the interval width. C) 15 is the numerator only. D) 17 is f(4). E) 8 mis-subtracts. Fix rule: average rate = output change over input change.
4. C. Correct: f(1) = −3, f'(1) = 2(1) − 4 = −2, so y + 3 = −2(x − 1) gives y = −2x − 1. A) flips the constant sign. B) uses the wrong slope. D) mis-distributes. E) uses f values incorrectly. Fix rule: point from f(a), slope from f'(a).
5. A. Correct: L(25.3) = 5 + (1/10)(0.3) = 5.03. B) 5.3 misplaces the decimal. C) 5.0299 is the true value, not the linear estimate. D) 5.15 uses slope 1/2. E) 5.006 mis-scales 0.3. Fix rule: L(x) = f(a) + f'(a)(x − a).
6. C. Correct: symmetric difference [g(6) − g(2)]/(6 − 2) = (17 − 5)/4 = 3. A) 6 uses a one-sided [g(6) − g(4)]/(6 − 4). B) 5 mis-subtracts. D) 2 divides by the wrong width. E) 3.5 averages the wrong pairs. Fix rule: center the interval on the point.
7. D. Correct: differentiable ⇒ continuous. A) a differentiable function need not pass through 0. B) and C) a nonzero slope is fine. E) differentiability at a does not force a second derivative. Fix rule: differentiability is the stronger condition and guarantees continuity.
8. D. Correct: √x has f''(x) = −1/(4x^{3/2}) < 0 (concave down), so the tangent line lies above the curve and 5.03 overestimates. A) and C) wrongly call the estimate low. B) the estimate is not exact. E) √x is not concave up. Fix rule: concave down → tangent above → overestimate.
9. B. Correct: |x − 3| is continuous at 3 (both pieces equal 0) but has a corner — left slope −1, right slope +1 — so it is not differentiable. A) and D) no single slope exists. C) the graph is unbroken. E) it is continuous. Fix rule: a corner is continuous but not differentiable.
10. D. Correct: [(x+h)² − x²]/h = 2x + h → 2x. A) 0 plugs in h = 0 prematurely. B) x halves the coefficient. C) x² is the function, not its rate. E) 2x + h forgets the limit. Fix rule: simplify the difference quotient, then let h → 0.
11. B. Correct: this is the limit definition of the derivative at a. A) and D) describe the difference quotient (a secant slope) before the limit. C) f(a) is a value. E) the limit exists whenever f is differentiable at a. Fix rule: the limit of the difference quotient is f'(a).
12. C. Correct: at a = 1, L(x) = ln 1 + 1·(x − 1) = x − 1, so L(1.1) = 0.1. A) 1.1 is the input. B) 0 is ln 1. D) 0.0953 is the true value, not the linear estimate. E) 1 is the slope. Fix rule: evaluate the tangent line, not the curve.