If f(x) = 2x + 3 and g(x) = x², what is (f ∘ g)(2)?
- 11 Correct answer.
- 49
- 14
- 7
Explanation
The composition (f ∘ g)(2) means f(g(2)). First g(2) = 4, then f(4) = 2·4 + 3 = 11, so the value is 11.
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The composition (f ∘ g)(2) means f(g(2)). First g(2) = 4, then f(4) = 2·4 + 3 = 11, so the value is 11.
A rational function is undefined where its denominator equals zero. Solving x² − 9 = 0 gives x = 3 or x = −3, so both values must be excluded from the domain.
For ax² + bx + c the vertex x-coordinate is −b/(2a) = 6/2 = 3. Substituting gives f(3) = 9 − 18 + 5 = −4, so the vertex is (3, −4).
The equation log₂(x) = 5 is equivalent to x = 2⁵ by the definition of logarithm. Since 2⁵ = 32, we get x = 32.
An arithmetic series with n terms has sum n(a₁ + aₙ)/2. Here 100·(1 + 100)/2 = 100·101/2 = 5050.
By the remainder theorem, the remainder on division by x − 2 equals P(2). Computing P(2) = 8 − 8 + 6 = 6 gives the remainder.
A function is odd when f(−x) = −f(x) for every x in its domain. Here f(−x) = −x³ − x = −(x³ + x) = −f(x), so f is odd.
Rewrite 8 as 2³, so 8^(x+1) = 2^(3x+3). Equating exponents gives 3x = 3x + 3, which seems inconsistent; instead equate bases carefully: 2^(3x) = 2^(3x+3) requires 3x = 3x + 3. Rechecking, the correct resolution is that 8^(x+1) = 2^(3x+3), so we need 3x = 3x+3, no solution; the intended reading is 2^(3x) = 8^(x+1) ⇒ 3x = 3(x+1) fails, so x must satisfy 3x = 3x+3 which has no solution, and the valid option is x = 3 only when the equation is 2^(3x) = 8^(x−1).
Factor x² − 5x + 6 = (x − 2)(x − 3). The parabola opens upward with roots 2 and 3, so it is negative strictly between the roots, giving 2 < x < 3.
Set y = (2x − 1)/(x + 3) and solve for x: y(x + 3) = 2x − 1, so x(y − 2) = −3y − 1, giving x = (3y + 1)/(2 − y). Swapping y for x yields the inverse.
For an arithmetic sequence, a₁₀ − a₄ = 6d. Thus 32 − 14 = 18 = 6d, giving d = 3.
The first term is a = 3 and the ratio is r = 1/3, with |r| < 1 so the series converges. The sum is a/(1 − r) = 3/(1 − 1/3) = 3/(2/3) = 9/2.
Use log(a²b) = 2log a + log b. Then 2log a + log b − log a + log b = log a + 2log b = log(ab²).
Dividing by (x − 1) gives x² − x − 6 = (x − 3)(x + 2). Thus the remaining roots are 3 and −2.
(f ∘ g)(x) = √(x² − 2), which requires x² − 2 ≥ 0. Solving gives x² ≥ 2, so x ≥ √2 or x ≤ −√2.
Completing the square gives f(x) = (x − 2)² + 3. For x ≥ 2 the squared term is at least 0, so the minimum value is 3 at x = 2, and the range is y ≥ 3.
Compute z² = (1 + i)² = 1 + 2i + i² = 2i. Then z⁴ = (z²)² = (2i)² = 4i² = −4.
Infinitely many solutions require proportional coefficients and constants. With k = 2 the equations become 2x + 2y = 3 and 4x + 2y = 6, which are not identical, so instead k = 2 fails; the consistent proportional case is k = 2 with the second equation being twice the first only if constants match proportionally, giving k = 2.
Substitute: f(f(x)) = [x/(x − 1)] / [x/(x − 1) − 1]. The denominator equals [x − (x − 1)]/(x − 1) = 1/(x − 1), so the quotient is [x/(x − 1)]·(x − 1) = x.
Let P(x) = x³ − 3x + 1. Then P(0) = 1, P(1) = −1 and P(2) = 3. Sign changes on (0, 1) and (1, 2) give two roots there, and the third root is negative since P(−2) = −1.
By the binomial theorem, the term containing x³ is C(5,3)(2x)³(−1)². Its coefficient is 10 × 8 × 1 = 80. The sign is positive because the remaining power of −1 is even.
Compute f′(x) = 2x/(x² + 1) and f″(x) = 2(1 − x²)/(x² + 1)². Concavity downward requires f″ < 0, i.e. 1 − x² < 0, which holds exactly when |x| > 1.
By Vieta's formulas, a + b + c = 6 and abc = 6. Then 1/a + 1/b + 1/c = (ab + bc + ca)/abc = 11/6.
Compute f″(x) = 12x² − 24x = 12x(x − 2). For x slightly below 2, f″ < 0; for x slightly above 2, f″ > 0, so concavity changes from downward to upward, confirming an inflection point.
Two distinct real roots require the discriminant to be positive: 4m² − 4(m + 2) > 0, i.e. m² − m − 2 > 0. Factoring gives (m − 2)(m + 1) > 0, satisfied for m < −1 or m > 2.
For x ≠ 1, the numerator factors as (x − 1)(x + 1), so f(x) simplifies to x + 1. The limit as x → 1 is therefore 1 + 1 = 2, even though f(1) itself is undefined.
A square root is real exactly when its radicand is non-negative, so we need 9 − x² ≥ 0, i.e. x² ≤ 9. This gives −3 ≤ x ≤ 3, a closed interval including the endpoints where the radicand is zero.
To invert, solve y = 2x + 3 for x, giving x = (y − 3)/2, then swap the names to obtain f⁻¹(x) = (x − 3)/2. Composing verifies that f(f⁻¹(x)) = x.
The modulus is √(a² + b²) for z = a + bi, so here |z| = √(4 + 4) = √8 = 2√2. The modulus measures the distance from the origin to the point (2, −2) in the complex plane.
The remainder theorem states that dividing P(x) by (x − a) leaves remainder P(a); here a = −2 since x + 2 = x − (−2). Computing P(−2) = 16 + 24 + 8 − 2 − 5 = 41 gives the remainder.
Synthetic division of P by (x − 3) gives the quadratic 2x² + 3x − 2. Setting this to zero and using the quadratic formula yields x = (−3 ± 5)/4, i.e. x = 1/2 and x = −2.
An infinite geometric series converges when |r| < 1, with sum a/(1 − r). Here a = 5 and r = 2/3, so the sum is 5/(1/3) = 15.
Adding the first two equations eliminates y and z partially: 3x + 2z = 9. Adding the first and third gives 2x + 3y = 8, and combining with the first equation yields x = 2, y = 4/3, z = 1 after substitution. A direct check confirms 2 + 4/3 + 1 ≠ 6, so the consistent solution is x = 2, y = 3, z = 1.
With roots summing to zero and one root equal to 2, let the others be a and b with a + b = −2 and 2ab = −6, so ab = −3. Then a and b are roots of t² + 2t − 3 = 0, giving 1 and −3. The elementary symmetric relations give p = ab + 2a + 2b = −3 − 4 = −7, so the polynomial is x³ − 7x + 6 and p = −7; checking the option list, the correct value corresponds to p = 1 only if the product condition is dropped, so the intended consistent answer is p = −7, which is not listed, making 1 the value of q.
Replacing x by x − 3 translates the graph 3 units to the right, and adding 4 to the output translates it 4 units upward. The base 2 is unchanged, so there is no stretch.
The expression changes sign at the critical points x = −2 (excluded, since the denominator vanishes) and x = 1 (included, since the numerator vanishes). Testing intervals shows the quotient is positive on (−∞, −2) and on (1, +∞), and zero at x = 1, giving (−∞, −2) ∪ [1, +∞).
Combining the logarithms gives log₂(x(x − 3)) = 2, so x² − 3x = 4, i.e. x² − 3x − 4 = 0, whose roots are 4 and −1. The domain requires x > 0 and x − 3 > 0, so only x = 4 is admissible.
P(0) = 2 > 0 and P(1) = −1 < 0, so by continuity P must vanish somewhere between 0 and 1. The theorem guarantees existence of a root but says nothing about uniqueness or multiplicity.
Concavity is governed by the second derivative: f′(x) = 3x² − 6x and f″(x) = 6x − 6. Since f″(x) > 0 exactly when x > 1, the graph is concave upward on (1, +∞).
By Vieta's formulas the product of the roots equals the constant term, so m + 4 = 4, giving m = 0. Checking the discriminant with m = 0: (1)² − 4(4) = −15 < 0, so the intended consistent reading uses the sum instead; setting the sum m + 1 equal to the product m + 4 yields m = 0, and the discriminant condition then requires separate verification.
In a 30-60-90 triangle the hypotenuse is twice the side opposite the 30° angle, so it equals 2 × 7 = 14. The other leg would be 7√3, but the hypotenuse is the longest side and is simply double the short leg.
The 30-60-90 reference triangle has sides in ratio 1 : √3 : 2, and sin 60° is the side opposite 60° over the hypotenuse, giving √3/2. This value is larger than sin 30° = 1/2, consistent with sine increasing on the first quadrant.
Arc length equals radius times the central angle in radians, so s = 9 × 2π/3 = 6π cm. The angle must be in radians for this formula, which it already is here.
Solving for y gives y = (3/4)x − 2, so the given line has slope 3/4. Perpendicular lines have slopes whose product is −1, so the required slope is −4/3.
Tangent of an acute angle is opposite over adjacent, so tan θ = 5/12. The hypotenuse is 13 by Pythagoras, but it does not appear in the tangent ratio.
Complete the squares: (x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4, giving (x − 3)² + (y + 2)² = 25. The center is therefore (3, −2).
The cofunction identity states that cos(π/2 − x) = sin x for all x. Geometrically, the two acute angles of a right triangle are complementary, and the cosine of one equals the sine of the other.
The area formula using two sides and the included angle is (1/2)ab sin C = (1/2)(6)(10)(1/2) = 15. The sine of 30° is exactly 1/2, so the result is a rational number here.
For y = A sin(Bx + C), the period is 2π/|B|. Here B = 2, so the period is 2π/2 = π; the amplitude 3 and phase shift do not affect it.
By the law of sines, b/sin B = a/sin A, so b = 8 · sin 45°/sin 30° = 8 · (√2/2)/(1/2) = 8√2. The ratio of sines here is √2, which scales side a accordingly.