⚡ Memory Point
LIMIT EXISTS ⟺ LEFT = RIGHT
\(\lim_{x\to a} f(x)\) exists only if \(\lim_{x\to a^-} = \lim_{x\to a^+}\)
Let \(f(x) = \begin{cases} x^2 + 1 & x < 2 \\ 3x - 1 & x \geq 2 \end{cases}\)
Does \(\displaystyle \lim_{x \to 2} f(x)\) exist?
Many students forget to check BOTH sides! Always check left AND right limit.
✅ Solution
Left: \(\lim_{x \to 2^-}(x^2+1) = 4+1 = 5\)
Right: \(\lim_{x \to 2^+}(3x-1) = 6-1 = 5\)
Since left = right = 5, the limit exists and equals 5. ✓
Q3 · Medium
🔵 Continuity
⚡ Memory Point
CONTINUITY = 3-CHECK RULE
① f(a) exists ② lim exists ③ f(a) = lim
📖 Think of it as:
"Can you draw the graph without lifting your pencil?" → Continuous ✓
Which value of \(k\) makes \(f(x) = \begin{cases} kx + 2 & x < 1 \\ x^2 + 3 & x \geq 1 \end{cases}\) continuous at \(x = 1\)?
✅ Solution
For continuity at \(x=1\): left limit = right limit
Right: \(1^2 + 3 = 4\)
Left: \(k(1) + 2 = k + 2\)
Set equal: \(k + 2 = 4 \Rightarrow k = \boxed{2}\)
② Derivatives
dy/dx
Q4 · Easy
🟢 Power Rule
⚡ Memory Point
POWER RULE: "BRING DOWN, REDUCE"
\(\frac{d}{dx}[x^n] = nx^{n-1}\) — bring the exponent down, then reduce it by 1
Find \(f'(x)\) if \(f(x) = 4x^3 - 6x^2 + 5x - 7\)
✅ Solution
Apply power rule term by term:
\(\frac{d}{dx}[4x^3] = 12x^2\)
\(\frac{d}{dx}[-6x^2] = -12x\)
\(\frac{d}{dx}[5x] = 5\), \(\frac{d}{dx}[-7] = 0\)
\(f'(x) = \boxed{12x^2 - 12x + 5}\)
Q5 · Medium
🟢 Chain Rule
⚡ Memory Point
CHAIN RULE: "OUTSIDE-INSIDE"
Derivative of outside (keep inside) × derivative of inside
⚡ Memory Point
PRODUCT RULE: "HI·D(LO) + LO·D(HI)"
\((fg)' = f'g + fg'\) — first·(deriv of second) + second·(deriv of first)
Find \(h'(x)\) if \(h(x) = x^2 \sin x\)
✅ Solution
\(f = x^2,\ f' = 2x\)
\(g = \sin x,\ g' = \cos x\)
\(h'(x) = f'g + fg' = 2x\sin x + x^2\cos x = \boxed{2x\sin x + x^2\cos x}\)
Q7 · Hard
🟢 Implicit Differentiation
⚡ Memory Point
IMPLICIT: "EVERY y gets ·(dy/dx)"
Differentiate both sides; multiply every dy/dx factor by dy/dx, then solve for dy/dx
Find \(\dfrac{dy}{dx}\) if \(x^2 + y^2 = 25\)
Students forget that \(\frac{d}{dx}[y^2] = 2y\cdot\frac{dy}{dx}\), NOT just 2y! The chain rule applies to y too.
✅ Solution
Differentiate both sides with respect to \(x\):
\(2x + 2y\dfrac{dy}{dx} = 0\)
Solve: \(2y\dfrac{dy}{dx} = -2x\)
\(\dfrac{dy}{dx} = \boxed{-\dfrac{x}{y}}\)
③ Applications of Derivatives
f'·f''
Q8 · Easy
🟠 Critical Points & Extrema
⚡ Memory Point
FIRST DERIVATIVE TEST: "+→−=MAX, −→+=MIN"
f' changes + to − → Local MAX. f' changes − to + → Local MIN
Find the critical points of \(f(x) = x^3 - 3x^2 - 9x + 5\)
📖 Key Step
Set \(f'(x) = 0\) and solve for \(x\). Critical points are where the slope = 0.
✅ Solution
\(f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)\)
Set \(f'(x) = 0\): \(x = 3\) or \(x = -1\)
Critical points: \(\boxed{x = -1 \text{ and } x = 3}\)
Q9 · Medium
🟠 Concavity & Inflection Points
⚡ Memory Point
SECOND DERIVATIVE TEST: "f''>0=SMILE (concave up), f''<0=FROWN (concave down)"
Inflection point: where f'' changes sign (concavity switches)
For \(f(x) = x^4 - 4x^3\), on what interval(s) is \(f\) concave up?
Inflection point ≠ where f''=0. You need to CHECK that f'' actually CHANGES SIGN!
✅ Solution
\(f'(x) = 4x^3 - 12x^2\)
\(f''(x) = 12x^2 - 24x = 12x(x - 2)\)
\(f''=0\) at \(x=0, x=2\)
Test: \(f''>0\) when \(x<0\) and \(x>2\) → Concave up on \(\boxed{(-\infty,0)\cup(2,\infty)}\)
Q10 · Hard
🟠 Related Rates
⚡ Memory Point
RELATED RATES: "DRAW → LABEL → DIFF w.r.t. TIME"
① Draw the situation ② Write an equation ③ Differentiate with respect to t (use chain rule!) ④ Plug in known values
A ladder 10 ft long leans against a wall. The bottom slides away at 2 ft/sec.
How fast is the top sliding down the wall when the bottom is 6 ft from the wall?
This is the most famous limit in calculus. Memorize it: the answer is always 1!
✅ Solution
Form: 0/0 → Apply L'Hôpital's Rule:
\(\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x} = \lim_{x\to 0}\dfrac{\cos x}{1} = \cos(0) = \boxed{1}\) This is also a fundamental trigonometric limit — memorize it!
Q19 · Hard
⭐ Optimization
⚡ Memory Point
OPTIMIZATION: "WRITE → REDUCE → DIFF → ZERO"
① Write objective function ② Use constraint to reduce to 1 variable ③ Differentiate ④ Set = 0
A farmer has 120 meters of fence and wants to enclose a rectangular area, using a barn wall as one side (no fence needed there). What width \(x\) maximizes the area?
📖 Setup
Constraint: \(2x + y = 120\), so \(y = 120 - 2x\)
Area: \(A = xy = x(120-2x)\)
✅ Solution
\(A(x) = x(120-2x) = 120x - 2x^2\)
\(A'(x) = 120 - 4x\)
Set \(A'(x) = 0\): \(120 = 4x \Rightarrow x = \boxed{30}\) m
\(A''(x) = -4 < 0\) → confirms maximum ✓
Max area = \(30 \times 60 = 1800\) m²
Q20 · Hard ★ FINAL BOSS
⭐ Accumulation Functions (Reading Graphs)
⚡ Memory Point
ACCUMULATION: "AREA ABOVE x-axis = +, BELOW = −"
g(x) = ∫f(t)dt. g increases when f > 0, decreases when f < 0, concave up when f is increasing
📖 Key Connections
If \(g(x) = \int_0^x f(t)\,dt\):
• \(g'(x) = f(x)\) (FTC Part 1)
• \(g\) is increasing where \(f(x) > 0\)
• \(g\) has a local max where \(f\) crosses from positive to negative
Let \(g(x) = \displaystyle\int_0^x f(t)\,dt\) where \(f\) is continuous.
Given that \(f(1)=3\), \(f(2)=0\), \(f(3)=-2\), and \(f\) is decreasing on \([0,4]\),
which statement is TRUE about \(g\)?
This is a conceptual question! You don't need to compute — use the relationship between g, g', and f.
✅ Solution
Since \(g'(x) = f(x)\):
• At \(x=2\): \(g'(2) = f(2) = 0\) → critical point
• For \(x<2\): \(f(x)>0\) → \(g' > 0\) → \(g\) increasing
• For \(x>2\): \(f(x)<0\) → \(g' < 0\) → \(g\) decreasing
→ g changes from increasing to decreasing at x=2 → Local MAXIMUM at x=2 ✓
Also: f is decreasing → g''=f'<0 → g is concave DOWN, not up.