Essential problems covering the topics students miss most. Pick an answer — instant feedback with full explanation.
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Functions & Graphs
01Domain & Range★☆☆
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KEY: Domain = what x CAN be · Range = what y CAN be · Square root → inside ≥ 0 · Fraction → bottom ≠ 0
What is the domain of the function below?
f(x) = √(x − 3)
Remember: you cannot take the square root of a negative number (in real numbers).
Explanation
For √(x − 3) to be defined, the expression inside must be ≥ 0.
So: x − 3 ≥ 0 → x ≥ 3.
Domain = [3, ∞). At x = 3 exactly, you get √0 = 0, which is fine!
02Composition★★☆
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KEY: (f ∘ g)(x) = f(g(x)) · Plug g(x) INTO f · Order matters! f∘g ≠ g∘f
If f(x) = 2x + 1 and g(x) = x², what is (f ∘ g)(3)?
Step: find g(3) first, then plug that result into f.
Explanation
Step 1: g(3) = 3² = 9
Step 2: f(9) = 2(9) + 1 = 19
Common mistake: computing f(3) = 7 first and then squaring → (g∘f)(3) = 49, not what we want!
03Inverse Functions★★☆
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KEY: Swap x ↔ y, then solve for y · f⁻¹ reflects over y = x line · f(f⁻¹(x)) = x always
What is the inverse of f(x) = 3x − 6?
Explanation
Replace f(x) with y: y = 3x − 6
Swap x and y: x = 3y − 6
Solve for y: x + 6 = 3y → y = (x + 6)/3
Check: f(f⁻¹(x)) = 3·(x+6)/3 − 6 = x + 6 − 6 = x ✓
This is a difference of squares: a² − b² = (a+b)(a−b)
Here a = x, b = 4, so: x² − 16 = (x+4)(x−4)
⚠️ Trap: (x−4)² = x² − 8x + 16, which is NOT the same!
05Polynomial Division★★☆
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KEY: Remainder Theorem → f(a) = remainder when divided by (x−a) · Faster than long division!
Find the remainder when p(x) = x³ − 2x + 5 is divided by (x − 2).
Hint: Use the Remainder Theorem — just evaluate p(2).
Explanation
Remainder Theorem: substitute x = 2 directly. p(2) = (2)³ − 2(2) + 5 = 8 − 4 + 5 = 9
If remainder = 0, then (x−2) would be a factor. Since it's 9 ≠ 0, it's not a factor.
06Zeros of Polynomials★★☆
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KEY: Zero = root = x-intercept = solution · Set f(x) = 0 · Factor first, then solve each factor
How many real zeros does f(x) = x³ − x have?
Explanation
Factor: x³ − x = x(x² − 1) = x(x+1)(x−1)
Set each factor = 0: x = 0, x = −1, x = 1
Three distinct real zeros. The graph crosses the x-axis at all three points.
Where is the vertical asymptote of the function below?
f(x) = (x + 2) / (x² − 9)
Explanation
Set denominator = 0: x² − 9 = 0 → (x+3)(x−3) = 0
So x = 3 and x = −3 are vertical asymptotes.
The numerator (x+2) ≠ 0 at these points, so no cancellation → both are true VAs, not holes.
08Horizontal Asymptote★★☆
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KEY: HA Rules → deg(top) < deg(bot): y=0 · deg(top) = deg(bot): y = leading coeffs · deg(top) > deg(bot): no HA
Find the horizontal asymptote of:
f(x) = (3x² − 1) / (x² + 5)
Explanation
Degrees of numerator and denominator are both 2 (equal degrees).
HA = ratio of leading coefficients = 3/1 = 3
So horizontal asymptote: y = 3
Use quotient rule: log₂(32) − log₂(4) = log₂(32/4) = log₂(8)
Now evaluate: 2³ = 8, so log₂(8) = 3
⚠️ Common mistake: subtracting the numbers inside → log₂(32−4) = log₂(28). WRONG!
10Exponential Equations★★☆
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KEY: Same base → set exponents equal · Different base → take log both sides · eˣ = k → x = ln(k)
Solve for x:
2^(x+1) = 16
Explanation
Rewrite 16 as a power of 2: 16 = 2⁴
Now: 2^(x+1) = 2⁴ → same base → exponents equal x + 1 = 4 → x = 3
11Natural Log★★☆
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KEY: ln and eˣ are inverses · ln(eˣ) = x always · e^(ln x) = x always · ln(1) = 0 · ln(e) = 1
Solve for x:
ln(x) + ln(x − 2) = ln(3)
Explanation
Combine: ln(x(x−2)) = ln(3)
So: x(x−2) = 3 → x² − 2x − 3 = 0 → (x−3)(x+1) = 0
x = 3 or x = −1. But check domain: ln(x) requires x > 0, and ln(x−2) requires x > 2.
x = −1 fails both → reject. Only x = 3 is valid.
Trigonometry
12Unit Circle★☆☆
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KEY: sin(θ) = y · cos(θ) = x · tan = sin/cos · ASTC: All Students Take Calculus (signs per quadrant)
What is the exact value of sin(5π/6)?
Think about which quadrant 5π/6 lies in and what the reference angle is.
Explanation
5π/6 is in Quadrant II (between π/2 and π). Reference angle = π − 5π/6 = π/6. sin(π/6) = 1/2
In Q II, sine is positive → sin(5π/6) = +1/2
⚠️ Don't confuse with cos(5π/6) = −√3/2 (negative in Q II).
KEY: Vertex form: y = a(x−h)² + k · Vertex = (h, k) · a > 0 opens up · a < 0 opens down · Axis of symmetry: x = h
What is the vertex of y = 2(x − 3)² − 5?
Explanation
In vertex form y = a(x−h)² + k, the vertex is (h, k).
Here h = 3 (note: it's x MINUS h, so the sign flips) and k = −5.
Vertex = (3, −5). ⚠️ Most common error: writing (−3, −5) by ignoring the sign rule.
17Circle Equation★☆☆
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KEY: (x−h)²+(y−k)² = r² · Center = (h,k) · Radius = r (not r²!) · Standard form must complete the square if needed
What is the center and radius of: (x − 2)² + (y + 3)² = 25?
Explanation
(x−h)²+(y−k)²=r² → here h=2, k=−3 (because it's y − (−3) = y + 3)
Center = (2, −3). Radius = √25 = 5 (not 25!).
⚠️ Common errors: forgetting to take √ for r, and getting the sign of k wrong.
18Systems of Equations★★☆
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KEY: Substitution: solve one variable, plug in · Elimination: add/subtract to cancel a variable · Check BOTH equations with your answer
Solve the system:
y = x + 1 x² + y² = 13
Explanation
Substitute y = x+1 into the circle: x² + (x+1)² = 13 x² + x² + 2x + 1 = 13 → 2x² + 2x − 12 = 0 → x² + x − 6 = 0
Factor: (x+3)(x−2) = 0 → x = −3 or x = 2
y = −3+1 = −2 and y = 2+1 = 3 → solutions: (2,3) and (−3,−2)
19Binomial Theorem★★★
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KEY: (a+b)ⁿ · Pascal's Triangle gives coefficients · nCr = n!/(r!(n−r)!) · Power of a decreases, b increases · Powers always sum to n
What is the coefficient of x² in the expansion of (x + 2)⁴?
Hint: Use the Binomial Theorem or Pascal's Triangle. The r-th term is C(4,r)·x^(4−r)·2^r.
Explanation
We want x²: exponent of x is 4−r = 2, so r = 2.
Term = C(4,2) · x² · 2² = 6 · x² · 4 = 24x²
Coefficient = 24.
⚠️ Common error: using C(4,2)=6 as the answer and forgetting to multiply by 2² = 4.
20Limits (Intro)★★☆
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KEY: If direct sub gives 0/0 → factor & cancel · Limit = value the function APPROACHES, not necessarily reaches · Cancel the problem factor first
Evaluate the limit:
lim(x→2) (x² − 4) / (x − 2)
Direct substitution gives 0/0. Factor the numerator first!
Explanation
Factor numerator: x² − 4 = (x+2)(x−2)
Cancel: (x+2)(x−2)/(x−2) = x+2 (for x ≠ 2)
Now take limit: lim(x→2)(x+2) = 2+2 = 4
The function has a hole at x=2, but the limit still exists and equals 4.