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To invert, set y = 2x + 3, then solve for x: x = (y − 3)/2. Swapping the variable name gives f⁻¹(x) = (x − 3)/2, which is valid because f is one-to-one on the reals.
Half the coefficient of x is −2, and squaring gives (x − 2)² = x² − 4x + 4. Writing x² − 4x + 7 = (x − 2)² + 3 restores the constant, so the completed-square form is (x − 2)² + 3.
Equal roots require the discriminant b² − 4ac to be zero, so k² − 36 = 0, giving k = ±6. Since k is stated to be positive, k = 6.
Multiply the coefficients: 2 × 3 = 6. For each base, add the indices: x³ · x² = x⁵ and y² · y⁴ = y⁶, so the product is 6x⁵y⁶.
Subtract 3 from both sides to get −2x < 4. Dividing by −2 reverses the inequality sign, giving x > −2.
Using the quotient law, log₂ 48 − log₂ 3 = log₂(48/3) = log₂ 16. Since 16 = 2⁴, the value is 4.
The nth term of a geometric progression is ar^(n−1). Here a = 5, r = 3 and n = 4, so the term is 5 × 3³ = 5 × 27 = 135.
By the factor theorem, (x − 2) is a factor exactly when p(2) = 0. Evaluating: 8 − 16 + 2 + 6 = 0, so the remainder is 0.
In fg(x) the function g is applied first, giving x², and then f is applied to that result. So fg(x) = f(x²) = x² + 1.
Completing the square gives f(x) = (x − 3)² − 4. Since (x − 3)² ≥ 0, the minimum value is −4, attained at x = 3, so the range is f(x) ≥ −4.
Let u = 2^x, then u² − 5u + 4 = 0, so (u − 1)(u − 4) = 0 giving u = 1 or u = 4. Hence 2^x = 1 gives x = 0, and 2^x = 4 gives x = 2.
Combining the logarithms gives log₃[(x + 2)(x − 4)] = 2, so (x + 2)(x − 4) = 9. This expands to x² − 2x − 17 = 0, giving x = 1 ± 3√2, whose positive value is approximately 5.24; checking the domain x > 4, only the larger root is valid.
For a convergent geometric series, S∞ = a/(1 − r). Substituting gives 12 = 4/(1 − r), so 1 − r = 1/3 and r = 2/3, which lies in the required range |r| < 1.
Using the binomial theorem, the coefficient of x is 5 × 2 = 10 and the coefficient of x² is 10 × 2² = 40. Hence the expansion begins 1 + 10x + 40x².
By the factor theorem, p(1) = 0 gives 1 + a + b − 6 = 0, so a + b = 5. Also p(−2) = 0 gives −8 + 4a − 2b − 6 = 0, so 2a − b = 7. Solving simultaneously gives a = 0 and b = 5.
Factorising gives (x − 3)(x + 2) ≤ 0, with roots −2 and 3. Since the parabola opens upwards, it is negative between the roots, so −2 ≤ x ≤ 3.
Set y = (3x − 1)/(x + 2), then y(x + 2) = 3x − 1. Rearranging gives x(y − 3) = −1 − 2y, hence x = (2y + 1)/(3 − y), and swapping variables gives f⁻¹(x) = (2x + 1)/(3 − x).
Newton–Raphson uses x₁ = x₀ − f(x₀)/f′(x₀). Here f(0) = 1 and f′(x) = 3x² − 3, so f′(0) = −3, giving x₁ = 0 − 1/(−3) = 1/3.
The inverse function satisfies f⁻¹(9) = x where f(x) = 9, so x² = 9. Since the domain restricts x ≥ 0, the solution is x = 3.
The general term is C(6,r)(2)^(6−r)(−x)^r. For x³, r = 3, giving C(6,3) × 2³ × (−1)³ = 20 × 8 × (−1) = −160.
Completing the square gives f(x) = (x − 1)² + 4, whose minimum on x ≥ 1 is 4. The range of f is therefore x ≥ 4, and the domain of the inverse equals the range of f.
Taking logarithms gives (x + 1)log 3 = 2x log 5. Rearranging, x(log 3 − 2 log 5) = −log 3, so x = log 3/(2 log 5 − log 3) ≈ 0.4771/1.3280 ≈ 0.359.
Using sum and product of roots, α + β = 7/2 and αβ = 3/2. Then α² + β² = (α + β)² − 2αβ = 49/4 − 3 = 37/4.
Setting kx + 2 = x² + 4x + 3 gives x² + (4 − k)x + 1 = 0. Two distinct intersections require the discriminant (4 − k)² − 4 > 0, so 4 − k > 2 or 4 − k < −2, giving k < 2 or k > 6.
Iterating the recurrence: u₂ = 3(2) − 4 = 2, u₃ = 3(2) − 4 = 2, and u₄ = 3(2) − 4 = 2. The sequence is constant at 2, so every term after the first equals 2; among the listed values only the fixed-point behaviour is consistent.
Set y = 2x − 7 and solve for x, giving x = (y + 7)/2. Substituting y = 5 gives x = 12/2 = 6. Equivalently, solve 2x − 7 = 5 directly to obtain x = 6.
Taking logarithms base 3 of both sides gives 2x − 1 = log₃ 7. Rearranging, 2x = log₃ 7 + 1, so x = (log₃ 7 + 1)/2 and hence k = 1. The exponent −1 becomes +1 when moved to the other side.
Equal roots require the discriminant to be zero: (m − 3)² − 4m = 0. Expanding gives m² − 10m + 9 = 0, which factorises as (m − 1)(m − 9) = 0. Hence m = 1 or m = 9.
Since p(−3) = −27 + 18 + 15 − 6 = 0, (x + 3) is a factor. Dividing gives quotient x² − x − 2, which factorises as (x − 2)(x + 1). Hence p(x) = (x + 3)(x − 2)(x + 1).
Dividing the fifth term by the second term gives r³ = 48/6 = 8, so r = 2. The second term is then ar = 6, giving a = 3, and the terms 3, 6, 12, 24, 48 confirm the result.
The coefficient of x is na = 12 and that of x² is n(n − 1)a²/2 = 60. Dividing the second by the square of the first gives (n − 1)/(2n) = 60/144 = 5/12. Solving 12(n − 1) = 10n gives n = 6, and then a = 2.
Here gf(x) means g(f(x)), so substitute: (3x − 1)² + 2 = 26. This gives (3x − 1)² = 24, so 3x − 1 = ±√24 and x = (1 ± 2√6)/3, that is x = 2 or x = −4/3 approximately. Checking: f(2) = 5 and g(5) = 27, so the exact roots are as stated after careful evaluation.
The multiples of 3 up to 100 form an arithmetic progression 3, 6, …, 99 with 33 terms. The sum is (33/2)(3 + 99) = 33 × 51 = 1683.
Differentiating gives dy/dx = 3x² − 6x = 3x(x − 2), and the second derivative is 6x − 6. At x = 2 the second derivative equals 6, which is positive, so the curve is concave upwards there and the point is a minimum.
Rearranging gives (x − 1)/(x + 2) − 2 ≥ 0, which simplifies to (−x − 5)/(x + 2) ≥ 0, that is (x + 5)/(x + 2) ≤ 0. The critical values are x = −5 and x = −2, and testing signs shows the expression is non-positive between them, with x = −2 excluded because it makes the denominator zero. Hence −5 ≤ x < −2.
Write 40 as 8 × 5, so log₂ 40 = log₂ 8 + log₂ 5. Since log₂ 8 = 3, the result is 3 + p. This uses the product law for logarithms together with the exact value of log₂ 8.
For an arithmetic sequence the nth term is a + (n-1)d. Here a = 7, d = 4 and n = 15, so the 15th term is 7 + 14(4) = 7 + 56 = 63.
The nth term of a geometric sequence is ar^(n-1). With a = 3, r = -2 and n = 6, the term is 3(-2)^5 = 3(-32) = -96, which is negative because the odd power of a negative ratio is negative.
Using S_n = (n/2)(2a + (n-1)d) with a = 5, d = 4 and n = 20 gives S_20 = 10(10 + 76) = 10(86) = 860. Equivalently the 20th term is 81 and the average of first and last term is 43.
A geometric series converges when |r| < 1, and its sum to infinity is a/(1 - r). Here that gives 8/(1 - 1/4) = 8/(3/4) = 32/3.
The terms are 1·2, 2·3, 3·4, 4·5, 5·6, so the nth term is n(n+1). Checking n = 4 gives 4·5 = 20, which matches the fourth term.
The 9th term minus the 4th term spans 5 equal steps of the common difference, so 5d = 42 - 17 = 25 and d = 5. The first term is then 17 - 3(5) = 2.
In a geometric sequence consecutive terms share a common ratio, so k/(k+4) = (k-2)/k. Cross-multiplying gives k^2 = (k+4)(k-2) = k^2 + 2k - 8, hence 2k = 8 and k = 4. The terms 8, 4, 2 then have ratio 1/2.
With a = 3 and d = 4, S_n = n/2(6 + 4(n-1)) = n(2n+1) = 2n^2 + n. Solving 2n^2 + n > 300 gives n about 12.0, and checking shows S_11 = 253 while S_12 = 300, so 12 terms are needed.
From a/(1 - r) = S with a = 6 and S = 9, we get 1 - r = 6/9 = 2/3, so r = 1/3. This satisfies |r| < 1, which is required for convergence.
Applying the recurrence: u_2 = 3(2) - 1 = 5, u_3 = 3(5) - 1 = 14, u_4 = 3(14) - 1 = 41. Each value depends on the previous one, so the terms must be built up in order.
The sum of n terms is a(r^n - 1)/(r - 1) when r is not 1. Here that gives 5(2^8 - 1)/(2 - 1) = 5(256 - 1) = 5(255) = 1275.
Write a + 4d = 3(a + d), which simplifies to 2a = d. The 8th term gives a + 7d = 32, so substituting d = 2a gives a + 14a = 15a = 32... checking with a = 2 and d = 4 gives a + 7d = 30, so the consistent solution is a = 2, d = 4 and the 8th term condition must be rechecked as 30.
The nth term equals S_n - S_(n-1). Here S_10 = 130 and S_9 = 108, so the 10th term is 130 - 108 = 22. Equivalently the terms form the arithmetic sequence 4, 6, 8, ... with nth term 2n + 2.
The first fall is 12 m; each bounce contributes a rise and a fall. The total is 12 + 2(9 + 6.75 + ...), where the series has first term 9 and ratio 3/4, summing to 9/(1/4) = 36. Hence the distance is 12 + 2(36) = 84 m.