GELE Mathematics — Differential CalculusRevision Notes
Revision notes for GELE Mathematics — Differential Calculus. Short, focused, and designed for the week before exam day. Use these when you are already familiar with the chapter and need a quick refresh on the high-yield items Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests.
Exam context
On the GELE 2026, the Mathematics subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Geodetic Engineering's pattern. Differential Calculus lands at position 5th out of 10 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Mathematics on a typical GELE paper.
Differential Calculus - Revision Notes
Differential calculus is one of the highest-yield topics in the MSTE (Mathematics, Surveying, and Transportation Engineering) portion of the PRC Civil Engineer Licensure Examination. Expect 8–12 items per board exam covering limits, differentiation rules, optimization (maxima-minima), and related rates. This chapter equips you to (1) evaluate limits including indeterminate forms, (2) apply all standard derivative rules fluently, (3) set up and solve optimization problems using a single-variable model, (4) differentiate implicit and composite functions, and (5) solve related-rates problems by differentiating geometric or physical relations with respect to time. Master the step-by-step strategies below and drill the worked examples until the procedures become automatic.
Sections
Formulas
Example
lim[x→0] (sin 3x)/(2x): Apply L'Hôpital — (3 cos 3x)/2 → 3/2 as x→0. Answer: 3/2
Formula
lim[x→a] f(x)/g(x) = lim[x→a] f'(x)/g'(x)
Variables
f and g are differentiable; g'(a) ≠ 0 after applying the rule; form must be 0/0 or ∞/∞
Application
Evaluate indeterminate-form limits that cannot be resolved by direct substitution or algebraic simplification
Example
lim[x→0] (sin 5x)/(3x) = (5/3) · lim[x→0] (sin 5x)/(5x) = 5/3
Formula
lim[x→0] (sin x)/x = 1
Variables
x in radians
Application
Trigonometric limits; often appears disguised as lim (sin kx)/(kx) = 1
Exam Tips
- Always attempt direct substitution first; only reach for L'Hôpital if an indeterminate form results.
- Factor and cancel common factors before differentiating — it is often faster than L'Hôpital.
- Memorize lim (sin x)/x = 1 and lim (1−cos x)/x = 0; they appear in disguised forms every exam.
- For limits at infinity of rational functions: divide every term by the highest power of x in the denominator.
Key Points
- The limit lim[x→a] f(x) = L means f(x) approaches L as x approaches a, regardless of f(a) itself.
- For a limit to exist, the left-hand limit and right-hand limit must be equal: lim[x→a⁻] f(x) = lim[x→a⁺] f(x).
- f is continuous at x = a if: (1) f(a) exists, (2) lim[x→a] f(x) exists, and (3) lim[x→a] f(x) = f(a).
- Indeterminate forms requiring further work: 0/0, ∞/∞, 0·∞, ∞−∞, 0⁰, 1^∞, ∞⁰.
- L'Hôpital's Rule: If the limit gives 0/0 or ∞/∞, differentiate numerator and denominator separately, then re-evaluate.
- Squeeze Theorem: If g(x) ≤ f(x) ≤ h(x) near a and lim g = lim h = L, then lim f = L.
- Standard limits to memorize: lim[x→0] (sin x)/x = 1; lim[x→0] (1 − cos x)/x = 0; lim[x→∞] (1 + 1/n)^n = e.
Definitions
Term
Limit
Definition
The value that f(x) approaches as x gets arbitrarily close to a, written lim[x→a] f(x) = L.
Importance
Foundation of all calculus; needed to define the derivative rigorously.
Term
Continuity
Definition
f is continuous at a if it is defined at a, the limit exists at a, and both values agree.
Importance
Differentiability requires continuity; discontinuous points are endpoints of optimization domains.
Term
Indeterminate Form
Definition
An expression whose limit cannot be determined by direct substitution because the result is 0/0, ∞/∞, etc.
Importance
Triggers the use of L'Hôpital's Rule or algebraic manipulation on board-exam problems.
Section Title
1. Limits and Continuity
Common Mistakes
- Applying L'Hôpital's Rule when the form is NOT indeterminate — always check the form first.
- Differentiating the whole fraction as a quotient instead of differentiating numerator and denominator separately.
- Using degrees instead of radians when evaluating trigonometric limits.
- Confusing 'limit does not exist' with 'limit equals infinity' — infinity is not a real number.
Formulas
Example
y = 4x³ − 2x⁻² + 7 → y' = 12x² + 4x⁻³ = 12x² + 4/x³
Formula
d/dx [xⁿ] = nxⁿ⁻¹
Variables
n = any real number
Application
Polynomial, radical, and negative-power functions
Example
y = x² sin x → y' = 2x sin x + x² cos x
Formula
(uv)' = u'v + uv'
Variables
u = u(x), v = v(x), both differentiable
Application
Product of two separate differentiable functions
Example
y = (x²+1)/(2x−3) → y' = [(2x)(2x−3) − (x²+1)(2)] / (2x−3)² = (2x²−6x−2)/(2x−3)²
Formula
(u/v)' = (u'v − uv') / v²
Variables
v ≠ 0
Application
Rational and trigonometric quotient functions
Example
y = (3x²+1)⁵ → y' = 5(3x²+1)⁴ · 6x = 30x(3x²+1)⁴
Formula
d/dx [f(g(x))] = f'(g(x)) · g'(x)
Variables
f = outer function, g = inner function
Application
Composite functions: powers of polynomials, trig of trig, exponentials with variable exponents
Example
y = sin(x³) → y' = cos(x³) · 3x²
Formula
d/dx [sin u] = cos u · u'; d/dx [cos u] = −sin u · u'
Variables
u = u(x)
Application
All trigonometric differentiation; chain rule automatically included
Example
y = e^(2x) → y' = 2e^(2x); y = ln(x²+1) → y' = 2x/(x²+1)
Formula
d/dx [eᵘ] = eᵘ · u'; d/dx [ln u] = u'/u
Variables
u = u(x), u > 0 for ln
Application
Exponential growth/decay models, logarithmic differentiation
Example
y = tan(3x) → y' = 3sec²(3x)
Formula
d/dx [tan u] = sec²u · u'; d/dx [sec u] = sec u tan u · u'
Variables
u = u(x)
Application
Tangent and secant differentiation in trig optimization problems
Exam Tips
- Board-exam composite functions like y = (2x+1)^10 are solved in seconds with the chain rule — never expand.
- Logarithmic differentiation simplifies products and quotients raised to powers: take ln of both sides first.
- For d/dx [aˣ] = aˣ ln a — do NOT write aˣ alone; the ln a factor is always present.
- When differentiating implicitly, group all dy/dx terms on the left, then factor and divide.
Key Points
- The derivative f'(x) gives the instantaneous rate of change and the slope of the tangent line at any point.
- Power Rule: d/dx [xⁿ] = nxⁿ⁻¹ — the single most frequently used rule on board exams.
- Product Rule: (uv)' = u'v + uv' — use whenever two functions are multiplied.
- Quotient Rule: (u/v)' = (u'v − uv')/v² — use whenever a function is divided by another.
- Chain Rule: d/dx [f(g(x))] = f'(g(x)) · g'(x) — the most commonly missed rule, especially in composite trig/exponential functions.
- Implicit differentiation: differentiate both sides w.r.t. x, treating y as a function of x; collect dy/dx terms.
- Higher-order derivatives: y'', y''' are successive derivatives; used in classifying critical points and curvature.
- Partial derivatives: ∂f/∂x treats all other variables as constants; ∂f/∂y treats x as constant.
Definitions
Term
Derivative
Definition
The derivative f'(x) = lim[h→0] [f(x+h) − f(x)]/h; it measures the instantaneous rate of change of f at x.
Importance
Core concept underpinning all differentiation; understanding this definition helps in deriving rules.
Term
Chain Rule
Definition
The rule for differentiating a function of a function: d/dx [f(g(x))] = f'(g(x)) · g'(x).
Importance
Most frequently omitted step on board exams — always identify the inner function and multiply by its derivative.
Term
Implicit Differentiation
Definition
Technique for finding dy/dx when y is not isolated; differentiate both sides w.r.t. x, applying the chain rule to y-terms.
Importance
Used in related-rates problems, tangent lines to curves like circles and ellipses.
Term
Partial Derivative
Definition
∂f/∂x: derivative of f(x,y) with respect to x, treating y as a constant.
Importance
Required for multivariable optimization and MSTE problems involving surfaces.
Section Title
2. Differentiation Rules and Techniques
Common Mistakes
- Forgetting to apply the chain rule on composite functions, e.g., writing d/dx [sin(x²)] = cos(x²) instead of 2x cos(x²).
- Sign errors in the quotient rule — memorize: 'lo d-hi minus hi d-lo, over lo-squared'.
- Differentiating constants as zero but failing to recognize that e³ is a constant (not e^(3x)).
- In implicit differentiation, forgetting to multiply by dy/dx when differentiating y-terms.
- Treating y' as a separate variable rather than dy/dx when collecting terms.
Formulas
Example
f(x) = −x² + 4x; f'(x) = −2x+4 = 0 → x=2; f''(2) = −2 < 0 → maximum. f(2) = 4
Formula
f'(x₀) = 0 and f''(x₀) < 0 ⟹ local maximum at x₀
Variables
x₀ = critical point; f' and f'' must exist
Application
Classify critical points without sign charts — faster for board exams
Example
Perimeter = 40 m → k=20; A = x(20−x); A' = 20−2x = 0 → x=10; A_max = 100 m²
Formula
A_rect = xy, constraint x+y = P/2 = k → maximize A = x(k−x) → x = k/2
Variables
x, y = dimensions; P = perimeter; k = semi-perimeter
Application
Maximum-area rectangle for a given perimeter; result: square is optimal
Example
30×30 sheet: V = x(30−2x)²; V' = (30−2x)² + x·2(30−2x)(−2) = 0 → 30−2x−4x = 0 → x=5 cm; V_max = 5(20)² = 2000 cm³
Formula
V_box = x(a−2x)(b−2x) where x = corner cut
Variables
a, b = sheet dimensions; x = height of box
Application
Open box optimization from rectangular sheet with square corners removed
Exam Tips
- Board-exam optimization problems always have a constraint equation — identify it first and use it to eliminate one variable.
- For symmetric shapes (square, cube, equilateral), the optimal solution is almost always the symmetric case — use this as a quick check.
- In fencing/perimeter problems along a river (one side free), the optimal rectangle has width = half the length.
- After solving x from f'(x) = 0, substitute back to find the optimized quantity — partial credit is lost if you stop at x.
- If the constraint gives two solutions, check both — one may be outside the physical domain.
Key Points
- A critical point occurs where f'(x) = 0 or f'(x) is undefined.
- First Derivative Test: f' changes + to − at x₀ → local maximum; f' changes − to + → local minimum.
- Second Derivative Test: f''(x₀) < 0 → local max; f''(x₀) > 0 → local min; f''(x₀) = 0 → inconclusive.
- Absolute (global) extrema on a closed interval [a, b]: evaluate f at all critical points AND endpoints; the largest/smallest value is the answer.
- Inflection point: f'' changes sign; curvature changes from concave up to concave down (or vice versa).
- Optimization strategy: identify the quantity to optimize, write it as a function of ONE variable using the constraint, differentiate, set equal to zero, solve, then verify.
- Common optimization shapes on board exams: rectangle of maximum area for fixed perimeter (square), cylinder of maximum volume, open box from a flat sheet.
Definitions
Term
Critical Point
Definition
A value x₀ in the domain of f where f'(x₀) = 0 or f'(x₀) is undefined.
Importance
All maxima and minima must occur at critical points or endpoints — the search begins here.
Term
Inflection Point
Definition
A point where the concavity of f changes; f''(x) = 0 is necessary but not sufficient.
Importance
Sometimes confused with extrema on board exams; inflection points are not necessarily extrema.
Term
Absolute vs. Local Extremum
Definition
A local extremum is the largest/smallest in a neighborhood; an absolute extremum is the largest/smallest on the entire domain.
Importance
On closed-interval problems, endpoints must be checked — a critical-point maximum may not be the global maximum.
Section Title
3. Maxima, Minima, and Optimization
Common Mistakes
- Forgetting to check endpoints when the problem specifies a closed interval or physical constraint (e.g., x > 0).
- Setting up the area/volume formula incorrectly before differentiating — always sketch the problem.
- Not verifying whether a critical point is a max or min — use either the first or second derivative test.
- Stopping after finding x without computing the maximum value of the quantity (the question often asks for the value, not x).
- Using the second-derivative test when f'' = 0 — switch to the first-derivative test in that case.
Formulas
Example
dV/dt = 10 m³/s, r = 2 m: dr/dt = 10/(4π·4) = 10/50.27 ≈ 0.199 m/s
Formula
dV/dt = 4πr² · dr/dt (sphere)
Variables
V = volume of sphere; r = radius; t = time
Application
Inflating balloons, expanding bubbles — find dr/dt given dV/dt or vice versa
Example
r = 5 m, dr/dt = 0.3 m/s: dA/dt = 2π(5)(0.3) = 3π ≈ 9.42 m²/s
Formula
dA/dt = 2πr · dr/dt (circle)
Variables
A = area of circle; r = radius
Application
Ripple problems, circular pool expanding
Example
L=5 m, x=3 m → y=4 m; dx/dt=0.5 m/s: dy/dt = −x(dx/dt)/y = −3(0.5)/4 = −0.375 m/s (top slides down)
Formula
2x·(dx/dt) + 2y·(dy/dt) = 0 (ladder/Pythagorean)
Variables
x = horizontal distance; y = height on wall; x² + y² = L² (L = fixed ladder length)
Application
Sliding ladder, receding boat, kite problems
Example
Conical tank, r/h = 1/2, so r = h/2. V = (π/12)h³; dV/dt = (π/4)h²·(dh/dt). At h=6m, dV/dt=−2m³/min: dh/dt = −2·4/(π·36) = −8/(36π) ≈ −0.0707 m/min
Formula
dV/dt = (1/3)π[r²(dh/dt) + h·2r(dr/dt)] (cone: general); dV/dt = (πr²/3)·dh/dt (similar-cone: r/h = constant)
Variables
r = base radius; h = height of cone
Application
Draining conical tanks — most frequent board-exam related-rates problem type
Exam Tips
- Always draw a diagram and assign variable names before writing any equation.
- In cone tank problems, write r in terms of h using r/h = R/H before differentiating — eliminates one variable.
- Label all given rates with their signs: filling = positive dV/dt; draining = negative dV/dt.
- Finish by checking units — if dV/dt is in m³/s and r is in m, then dr/dt should be in m/s.
- Ladder problems: once x and dx/dt are given, find y from the Pythagorean theorem before substituting.
Key Points
- Related rates problems involve two or more quantities that both change with time; you differentiate their relationship with respect to t.
- Step 1: Draw and label a diagram identifying all time-varying quantities.
- Step 2: Write an equation relating the variables (geometric formula, Pythagorean theorem, similar triangles, etc.).
- Step 3: Differentiate both sides with respect to t using the chain rule.
- Step 4: Substitute the given numerical values (including the given rates) AFTER differentiating — NEVER before.
- Step 5: Solve for the unknown rate.
- Common geometric relations: sphere V = (4/3)πr³; circle A = πr²; cone V = (1/3)πr²h; right triangle: x² + y² = L².
- Rates are derivatives: dV/dt = rate of volume change; dr/dt = rate of radius change; dx/dt = speed of a moving point.
Definitions
Term
Rate of Change
Definition
The derivative of a quantity with respect to time, e.g., dV/dt = rate of volume change in m³/s.
Importance
Related rates problems are entirely about connecting rates through a differentiating equation.
Term
Similar Triangle Proportion
Definition
In cone problems, the ratio r/h = R/H (constant) allows expressing r in terms of h alone, reducing to one variable.
Importance
Without this substitution, cone problems require implicit differentiation of two unknowns — use the proportion.
Section Title
4. Related Rates
Common Mistakes
- Substituting numerical values BEFORE differentiating — the most common and costly error in related-rates problems.
- Getting the sign of rates wrong — decreasing quantities have negative rates; always assign signs consistently.
- Not using the similar-triangle ratio in cone problems, leading to two unknowns after differentiation.
- Confusing 'the shadow lengthening' rate with 'the tip of the shadow moving' rate in lamppost problems — these are different quantities.
- Forgetting the chain rule factor (dr/dt or dy/dt) when differentiating geometric formulas with respect to t.
Formulas
Example
y = x³ at x₀=1: y₀=1, y'=3x²→3; tangent: y−1=3(x−1) → y = 3x−2
Formula
y − y₀ = f'(x₀)(x − x₀)
Variables
x₀ = point of tangency; y₀ = f(x₀); f'(x₀) = slope
Application
Finding tangent line equation to any differentiable curve
Example
y = x² at origin: y'(0)=0, y''(0)=2; R = [1+0]^(3/2)/2 = 1/2
Formula
R = [1 + (y')²]^(3/2) / |y''|
Variables
y' = first derivative; y'' = second derivative; R = radius of curvature
Application
Road/rail alignment design; finding the tightest bend of a curve
Example
f(x)=x² on [1,3]: MVT gives 2c = (9−1)/(3−1) = 4 → c = 2 ∈ (1,3) ✓
Formula
f'(c) = [f(b) − f(a)] / (b − a)
Variables
c ∈ (a, b); f continuous on [a,b] and differentiable on (a,b)
Application
Mean Value Theorem; proving existence of specific derivative values
Exam Tips
- For radius of curvature at the origin or vertex of a parabola, y'=0, so R simplifies to 1/|y''| — very fast to compute.
- Normal line slope is −1/f'(x₀); if f'(x₀)=0 (horizontal tangent), the normal is vertical (undefined slope — write x = x₀).
- Curve analysis: always find critical points AND inflection points before sketching — board multiple-choice options differ by concavity.
- MVT problems: set f'(x) = average rate, solve for x, and verify x is inside the open interval.
Key Points
- Tangent line at (x₀, y₀): slope m = f'(x₀); equation: y − y₀ = m(x − x₀).
- Normal line: perpendicular to tangent; slope = −1/m.
- Radius of curvature R = [1 + (y')²]^(3/2) / |y''|; large R means nearly flat, small R means sharply curved.
- Center of curvature (evolute): the center of the osculating circle at a point.
- Curve sketching sequence: domain → intercepts → symmetry → asymptotes → critical points → inflection points → sketch.
- Rolle's Theorem: If f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then ∃ c ∈ (a,b) where f'(c) = 0.
- Mean Value Theorem (MVT): ∃ c ∈ (a,b) such that f'(c) = [f(b)−f(a)]/(b−a) — the derivative equals the average rate of change.
Definitions
Term
Radius of Curvature
Definition
R = [1+(y')²]^(3/2)/|y''|; the radius of the circle that best fits the curve at a given point.
Importance
Appears in engineering design (road curves, beam deflection) and board-exam problems directly.
Term
Inflection Point
Definition
A point where y'' = 0 AND y'' changes sign — the curve changes concavity.
Importance
Distinguish from critical points: inflection points are not extrema unless f' = 0 also holds.
Term
Mean Value Theorem
Definition
Guarantees a point c where the instantaneous slope equals the average slope over an interval.
Importance
Foundation for many calculus proofs; occasionally tested directly on the licensure exam.
Section Title
5. Tangent Lines, Curvature, and Other Applications
Common Mistakes
- Confusing tangent slope with normal slope — they are negative reciprocals, not equal.
- Applying R = [1+(y')²]^(3/2)/|y''| with y' and y'' evaluated at different x values.
- Declaring an inflection point at x where y''=0 without verifying a sign change in y''.
- Mixing up Rolle's Theorem (requires f(a)=f(b)) with the general MVT.
Connections
- Differential Calculus → Integral Calculus: The Fundamental Theorem of Calculus links differentiation and integration — derivatives of integrals and antiderivatives of functions are board-exam bridge topics.
- Differential Calculus → Structural Analysis: Slope and deflection of beams involve EI·y'' = M(x); the bending moment diagram is derived by integrating the load function and the slope by differentiating the deflection.
- Differential Calculus → Fluid Mechanics: Rate-of-flow problems (dV/dt, dQ/dt) use the same related-rates framework applied to pipe flow and storage tanks.
- Differential Calculus → Physics / Engineering Mechanics: velocity = ds/dt (first derivative of displacement), acceleration = d²s/dt² (second derivative) — directly used in kinematics problems on the board exam.
- Differential Calculus → Optimization in Engineering Design: Minimum cost/maximum capacity problems in surveying (route alignment), geotechnical (earth volume), and structural engineering (minimum steel area) all reduce to f'(x)=0.
- Maxima-Minima → NSCP 2015 Load Combinations: While not calculus per se, optimization of load effects and structural efficiency uses the concept of extreme values.
- Differential Calculus → Numerical Methods: Newton-Raphson method for root-finding uses f'(x) iteratively — a direct application of differentiation reviewed in the Engineering Sciences portion.
Exam Strategy
For the MSTE board-exam differential calculus section, allocate time as follows: (1) Limits — 1 min each; direct substitution first, then L'Hôpital if needed. (2) Differentiation — 1.5 min each; identify the rule type (power, product, quotient, chain) before writing anything. (3) Maxima-minima — 3–4 min each; spend the first 30 seconds drawing the diagram and labeling variables, then set up the constraint and objective function before differentiating. (4) Related rates — 3–4 min each; write the geometric equation, differentiate w.r.t. t, THEN substitute numbers. High-yield topics based on past board exams: (a) Optimization of rectangles, boxes, and cylinders — appears almost every exam; (b) Sliding-ladder and conical-tank related rates — classic recurring items; (c) Chain-rule composite functions — 2–3 items every exam; (d) Radius of curvature — 1 item, very formula-direct. Use the process of elimination on multiple-choice: if the problem has a symmetric constraint (fixed perimeter, fixed surface area), the answer is almost always the most symmetric shape (square, cube, hemisphere). Always verify your critical point using the second-derivative test before selecting the final answer.
Quick Review Questions
Evaluate: lim[x→0] (tan 3x) / (sin 5x)
Both numerator and denominator → 0 as x→0 (form 0/0). Use small-angle approximations: tan 3x ≈ 3x and sin 5x ≈ 5x near zero, so the limit = 3x/5x = 3/5. Alternatively, L'Hôpital: (3sec²3x)/(5cos5x) → 3(1)/5(1) = 3/5.
Find y' for y = (x² + 3)⁴ · e^(2x).
Product rule: u=(x²+3)⁴, v=e^(2x). u'=4(x²+3)³·2x=8x(x²+3)³; v'=2e^(2x). y' = 8x(x²+3)³·e^(2x) + (x²+3)⁴·2e^(2x) = 2e^(2x)(x²+3)³[4x + (x²+3)].
A farmer has 120 m of fencing to enclose a rectangular area with one side against a barn (no fencing needed on that side). Find the dimensions for maximum area.
Let width = x (two sides), length = y (one side free). Constraint: 2x + y = 120 → y = 120−2x. A = xy = x(120−2x). dA/dx = 120−4x = 0 → x=30 m; y=60 m. A_max = 30×60 = 1800 m². Check: d²A/dx²=−4<0 ✓ (maximum).
The radius of a circle is increasing at 2 cm/s. How fast is the area increasing when r = 5 cm?
A = πr². Differentiate w.r.t. t: dA/dt = 2πr·(dr/dt). Substitute r=5, dr/dt=2: dA/dt = 2π(5)(2) = 20π cm²/s.
For y = 3x³ − 5x² + 2x, find y'(2).
y' = 9x² − 10x + 2. At x=2: y'(2) = 9(4) − 10(2) + 2 = 36 − 20 + 2 = 18.
Find the radius of curvature of y = x² at the origin.
y' = 2x → y'(0) = 0. y'' = 2 → y''(0) = 2. R = [1+(y')²]^(3/2)/|y''| = [1+0]^(3/2)/2 = 1/2 = 0.5.
A 5-m ladder leans against a wall. The base slides away at 0.5 m/s. How fast is the top sliding down when the base is 3 m from the wall?
x²+y²=25. At x=3: y=√(25−9)=4 m. Differentiate: 2x(dx/dt)+2y(dy/dt)=0. Substitute x=3, y=4, dx/dt=0.5: 2(3)(0.5)+2(4)(dy/dt)=0 → 3+8(dy/dt)=0 → dy/dt=−3/8=−0.375 m/s.
Find all critical points of f(x) = x³ − 3x² − 9x + 5 and classify them.
f'(x) = 3x²−6x−9 = 3(x²−2x−3) = 3(x−3)(x+1) = 0 → x=3 or x=−1. f''(x)=6x−6. f''(−1)=−12<0 → local max; f''(3)=12>0 → local min. f(−1)=(−1)³−3(1)−9(−1)+5=−1−3+9+5=10; f(3)=27−27−27+5=−22.
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