Topic Review — Introduction to Differential Calculus — Solutions
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Fluency
Q1 — Average rate of change
For f(x) = x² + 3x, find the average rate of change over each interval.
(a) from x = 1 to x = 3 (b) from x = −2 to x = 0 (c) from x = 2 to x = 2 + h (leave in terms of h)
(a) f(1) = 1 + 3 = 4, f(3) = 9 + 9 = 18. Average rate = (18 − 4)/(3 − 1) = 14/2 = 7
(b) f(−2) = 4 − 6 = −2, f(0) = 0. Average rate = (0 − (−2))/(0 − (−2)) = 2/2 = 1
(c) f(2 + h) = (2 + h)² + 3(2 + h) = 4 + 4h + h² + 6 + 3h = h² + 7h + 10. f(2) = 4 + 6 = 10. Average rate = (h² + 7h + 10 − 10)/h = (h² + 7h)/h = h + 7
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Fluency
Q2 — Evaluating limits
Evaluate each limit.
(a) limx→3 (x² − 9)/(x − 3) (b) limh→0 (3h² + 5h)/h (c) limh→0 (h² + 4h)/h
(a) Factor: (x² − 9)/(x − 3) = (x − 3)(x + 3)/(x − 3) = x + 3. limx→3 (x + 3) = 6
(b) (3h² + 5h)/h = 3h + 5. limh→0 (3h + 5) = 5
(c) (h² + 4h)/h = h + 4. limh→0 (h + 4) = 4
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Fluency
Q3 — First principles for a constant and linear function
Use the definition f′(x) = limh→0 [f(x + h) − f(x)]/h to differentiate:
(a) f(x) = 7 (b) f(x) = 4x − 1 (c) f(x) = 3x
(a) f(x + h) = 7. [f(x + h) − f(x)]/h = (7 − 7)/h = 0. limh→0 0 = f′(x) = 0 (constant rule)
(b) f(x + h) = 4(x + h) − 1 = 4x + 4h − 1. [f(x + h) − f(x)]/h = (4x + 4h − 1 − 4x + 1)/h = 4h/h = 4. f′(x) = 4
(c) [3(x + h) − 3x]/h = 3h/h = 3. f′(x) = 3
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Fluency
Q4 — First principles for x²
Using first principles, prove that d/dx[x²] = 2x. Show all steps clearly.
f(x) = x², f(x + h) = (x + h)² = x² + 2xh + h²
[f(x + h) − f(x)]/h = (x² + 2xh + h² − x²)/h = (2xh + h²)/h = 2x + h
limh→0 (2x + h) = f′(x) = 2x ✓
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Fluency
Q5 — Power rule: differentiating monomials
Differentiate each using the power rule d/dx[xn] = nxn−1.
(a) y = x5 (b) y = x8 (c) y = x1/2 (d) y = x−3 (e) y = 6x4
(a) 5x4
(b) 8x7
(c) ½x−1/2 = 1/(2√x)
(d) −3x−4 = −3/x4
(e) 24x3
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Fluency
Q6 — Differentiating polynomials
Find f′(x) for each polynomial.
(a) f(x) = x3 + 4x² − 7x + 2 (b) f(x) = 5x4 − 3x² + 1 (c) f(x) = 2x3 − x + 9
(a) f′(x) = 3x² + 8x − 7
(b) f′(x) = 20x3 − 6x
(c) f′(x) = 6x² − 1
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Understanding
Q7 — First principles for x³
Using first principles, show that d/dx[x³] = 3x². Expand (x + h)³ = x³ + 3x²h + 3xh² + h³ fully.
f(x + h) − f(x) = (x + h)³ − x³ = x³ + 3x²h + 3xh² + h³ − x³ = 3x²h + 3xh² + h³
[f(x + h) − f(x)]/h = (3x²h + 3xh² + h³)/h = 3x² + 3xh + h²
limh→0 (3x² + 3xh + h²) = 3x² ✓
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Understanding
Q8 — Evaluating the derivative at a point
Find the gradient of each curve at the specified x-value.
(a) f(x) = x³ − 2x at x = 2 (b) f(x) = 4x² + 1 at x = −1 (c) f(x) = x4 − x at x = 1
(a) f′(x) = 3x² − 2. f′(2) = 12 − 2 = 10
(b) f′(x) = 8x. f′(−1) = −8
(c) f′(x) = 4x³ − 1. f′(1) = 4 − 1 = 3
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Understanding
Q9 — Interpreting the derivative as gradient
For the curve y = 3x² − 12x + 5:
(a) Find dy/dx. (b) Find the gradient at x = 1 and at x = 3. (c) What do your answers tell you about the curve at those points?
(a) dy/dx = 6x − 12
(b) At x = 1: gradient = 6 − 12 = −6. At x = 3: gradient = 18 − 12 = 6
(c) At x = 1 the curve is decreasing (negative gradient). At x = 3 the curve is increasing (positive gradient). The curve has a turning point between these values.
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Understanding
Q10 — Finding where the derivative equals a given value
Find the x-value(s) where the gradient of the curve equals the given value.
(a) y = x² − 4x + 1, gradient = 0 (b) y = 2x³ − 3x, gradient = 6 (c) y = x³ − 12x, gradient = 0
(a) dy/dx = 2x − 4. Set 2x − 4 = 0 → x = 2
(b) dy/dx = 6x² − 3. Set 6x² − 3 = 6 → 6x² = 9 → x² = 3/2 → x = ±√(3/2) = ±√6/2
(c) dy/dx = 3x² − 12. Set 3x² − 12 = 0 → x² = 4 → x = ±2
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Understanding
Q11 — Derivative notation
Let f(x) = 2x4 − 5x² + 3x − 1.
(a) Find f′(x). (b) Evaluate f′(0) and f′(2). (c) Write the derivative using dy/dx notation if y = f(x).
(a) f′(x) = 8x³ − 10x + 3
(b) f′(0) = 0 − 0 + 3 = 3. f′(2) = 8(8) − 10(2) + 3 = 64 − 20 + 3 = 47
(c) dy/dx = 8x³ − 10x + 3
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Understanding
Q12 — First principles from the definition
Use first principles to differentiate f(x) = x² − 5x.
Show that your result matches the power rule applied term by term.
f(x + h) = (x + h)² − 5(x + h) = x² + 2xh + h² − 5x − 5h
f(x + h) − f(x) = 2xh + h² − 5h
[f(x + h) − f(x)]/h = 2x + h − 5
limh→0 (2x + h − 5) = f′(x) = 2x − 5
Power rule: d/dx[x²] = 2x, d/dx[−5x] = −5. Sum: 2x − 5 ✓
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Problem Solving
Q13 — The derivative function
For the function f(x) = x³ − 3x:
(a) Find f′(x). (b) Sketch a sign diagram for f′(x) to identify where f is increasing and decreasing. (c) State the x-coordinates of any turning points of f.
(a) f′(x) = 3x² − 3 = 3(x² − 1) = 3(x − 1)(x + 1)
(b) f′(x) = 0 at x = −1 and x = 1. Sign diagram: f′ > 0 for x < −1, f′ < 0 for −1 < x < 1, f′ > 0 for x > 1. So f is increasing, then decreasing, then increasing.
(c) Turning points at x = −1 (local maximum) and x = 1 (local minimum).
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Problem Solving
Q14 — Average rate from first principles limit
The height (in metres) of a ball thrown upward is h(t) = −5t² + 20t + 1, where t is time in seconds.
(a) Find the average rate of change of height from t = 1 to t = 3. (b) Using first principles, find h′(t). (c) Find the instantaneous velocity at t = 1 and t = 3. (d) When is the ball at its highest point?
(a) h(1) = −5 + 20 + 1 = 16. h(3) = −45 + 60 + 1 = 16. Average rate = (16 − 16)/(3 − 1) = 0 m/s
(b) h(t + k) = −5(t + k)² + 20(t + k) + 1 = −5t² − 10tk − 5k² + 20t + 20k + 1. h(t + k) − h(t) = −10tk − 5k² + 20k. Divide by k: −10t − 5k + 20. limk→0: h′(t) = −10t + 20
(c) h′(1) = −10 + 20 = 10 m/s (moving upward). h′(3) = −30 + 20 = −10 m/s (moving downward)
(d) At highest point h′(t) = 0: −10t + 20 = 0 → t = 2 seconds
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Problem Solving
Q15 — Extended problem: connecting concepts
Two curves are defined by f(x) = x3 − 6x and g(x) = −x² + 2.
(a) Find f′(x) and g′(x). (b) At what x-value do the two curves have the same gradient? (c) At that x-value, are both curves increasing, decreasing, or one of each? (d) Find the equation of the tangent to f at this x-value.
(a) f′(x) = 3x² − 6 and g′(x) = −2x
(b) Set f′(x) = g′(x): 3x² − 6 = −2x → 3x² + 2x − 6 = 0. Using quadratic formula: x = (−2 ± √(4 + 72))/6 = (−2 ± √76)/6 = (−1 ± √19)/3. So x = (−1 + √19)/3 ≈ 1.12 or x = (−1 − √19)/3 ≈ −1.79
(c) At x ≈ 1.12: f′(1.12) = g′(1.12) ≈ −2(1.12) ≈ −2.24 < 0. Both curves are decreasing at this point.
(d) At x = (−1 + √19)/3: f(x) = x³ − 6x. Gradient m = −2x ≈ −2.24. Tangent: y − f(x) = m(x − x0). [Exact answer left in terms of the exact x-value; a numerical answer at x ≈ 1.12 is acceptable: y ≈ −2.24(x − 1.12) + f(1.12) where f(1.12) ≈ 1.40 − 6.72 ≈ −5.32, so y ≈ −2.24x − 2.81]