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To find the inverse, set y = 2x + 3 and solve for x: x = (y - 3)/2. Substituting y = 7 gives x = (7 - 3)/2 = 2. Therefore f^{-1}(7) = 2.
The notation fg(x) means f(g(x)). Substitute g(x) into f: f(3x + 1) = (3x + 1)^2 - 4 = 9x^2 + 6x + 1 - 4 = 9x^2 + 6x - 3.
Divide P(x) by (x - 1) using synthetic division or long division to obtain x^2 - x - 6. Factorising x^2 - x - 6 gives (x - 3)(x + 2). Hence the other factors are (x + 2) and (x - 3).
Let y = 2^x. The equation becomes y^2 - 5y + 4 = 0, which factorises to (y - 1)(y - 4) = 0. So y = 1 or y = 4, giving 2^x = 1 (x = 0) or 2^x = 4 (x = 2).
Using the laws of logarithms, log_a(8) + log_a(4) = log_a(32), and log_a(32) - log_a(2) = log_a(16). Alternatively, combine as log_a(8×4/2) = log_a(16).
For equal roots the discriminant b^2 - 4ac must be zero. Here a = 1, b = k, c = 9, so k^2 - 36 = 0, giving k = ±6.
Complete the square: f(x) = 2(x^2 - 4x) + 5 = 2[(x - 2)^2 - 4] + 5 = 2(x - 2)^2 - 3. The vertex is therefore (2, -3).
Set y = (x + 1)/(x - 2). To find f^{-1}(3), solve 3 = (x + 1)/(x - 2). Cross-multiplying gives 3x - 6 = x + 1, so 2x = 7 and x = 7/2.
Factor out 3 from the first two terms: 3(x^2 - 4x) + 7. Complete the square inside: x^2 - 4x = (x - 2)^2 - 4. Thus 3[(x - 2)^2 - 4] + 7 = 3(x - 2)^2 - 12 + 7 = 3(x - 2)^2 - 5.
Compute fg(x) = f(x^2 + 1) = 3(x^2 + 1) - 2 = 3x^2 + 1. Compute gf(x) = g(3x - 2) = (3x - 2)^2 + 1 = 9x^2 - 12x + 5. Setting equal gives 3x^2 + 1 = 9x^2 - 12x + 5, so 6x^2 - 12x + 4 = 0, or 3x^2 - 6x + 2 = 0. This does not simplify to x = 1/2; instead, solving gives x = 1 ± √3/3. Wait, check: 3x^2 - 6x + 2 = 0 gives x = [6 ± √(36 - 24)]/6 = [6 ± √12]/6 = 1 ± √3/3. None of the options match. Let me recompute: fg(x) = f(g(x)) = 3(x^2 + 1) - 2 = 3x^2 + 1. gf(x) = g(f(x)) = (3x - 2)^2 + 1 = 9x^2 - 12x + 4 + 1 = 9x^2 - 12x + 5. Equate: 3x^2 + 1 = 9x^2 - 12x + 5 → 6x^2 - 12x + 4 = 0 → 3x^2 - 6x + 2 = 0. Discriminant 36 - 24 = 12, roots (6 ± 2√3)/6 = 1 ± √3/3. None of the given options match. So this question is flawed. I need to replace it with a valid question. Let's design a different one: maybe f(x)=2x+1, g(x)=x^2, solve fg(x)=gf(x). fg(x)=2x^2+1, gf(x)=(2x+1)^2=4x^2+4x+1. Equate: 2x^2+1=4x^2+4x+1 → 2x^2+4x=0 → 2x(x+2)=0 → x=0 or x=-2. Options could include 0 and -2. But need one correct. Maybe ask for the value of x that satisfies, with options including 0. But then two solutions. Better to ask: The functions f and g are defined by f(x)=2x+1 and g(x)=x^2. Find the values of x for which fg(x)=gf(x). Options: x=0 or x=-2; x=0 or x=2; x=1 or x=-1; x=0 only. Correct: x=0 or x=-2. That works. Let's use that.
Find the second derivative: y' = 3x^2 - 6x, y'' = 6x - 6. At x = 0, y'' = -6 < 0, so it is a maximum. At x = 2, y'' = 6 > 0, so it is a minimum.
The critical values are x = 1 (numerator zero) and x = -2 (denominator zero, excluded). Testing intervals: for x < -2, both numerator and denominator are negative, so the fraction is positive; for -2 < x < 1, numerator negative and denominator positive, fraction negative; for x > 1, both positive, fraction positive. At x = 1 the fraction is zero, which satisfies ≤ 0. Thus the solution is -2 < x ≤ 1.
Complete the square: f(x) = (x + 2)^2 + 3. Since the domain is x ≥ -2, the range is f(x) ≥ 3. Let y = (x + 2)^2 + 3, so (x + 2)^2 = y - 3. Taking the positive square root because x + 2 ≥ 0, we get x + 2 = √(y - 3), hence x = -2 + √(y - 3). Thus f^{-1}(x) = -2 + √(x - 3).
For two distinct real roots, the discriminant must be positive: (k - 2)^2 - 4(1)(k + 1) > 0. Expanding gives k^2 - 4k + 4 - 4k - 4 > 0, so k^2 - 8k > 0, or k(k - 8) > 0. This inequality holds when k < 0 or k > 8.
Since (x - 1) is a factor, P(1) = 0: 2 + a + b - 6 = 0 → a + b = 4. Since (x + 2) is a factor, P(-2) = 0: -16 + 4a - 2b - 6 = 0 → 4a - 2b = 22 → 2a - b = 11. Solving the simultaneous equations a + b = 4 and 2a - b = 11 gives 3a = 15, so a = 5 and b = -1? Wait, check: a + b = 4, 2a - b = 11. Add: 3a = 15 → a = 5, then b = -1. But options don't include a=5, b=-1. Let me recompute P(-2): P(-2) = 2(-8) + a(4) + b(-2) - 6 = -16 + 4a - 2b - 6 = 4a - 2b - 22 = 0 → 2a - b = 11. Correct. Then a + b = 4. Adding gives 3a = 15 → a = 5, b = -1. This is not in options. So my question is flawed. I need to adjust the polynomial or the factors. Let's choose a polynomial that yields one of the options. Suppose P(x) = 2x^3 + ax^2 + bx - 6 has factors (x - 1) and (x + 2). Let's find a and b correctly: a=5, b=-1. Not in options. So I will change the options to include a=5, b=-1. But then need to ensure only one correct. Let's rewrite the question with different factors or polynomial. Maybe P(x) = x^3 + ax^2 + bx - 6 with factors (x - 1) and (x + 2). Then P(1)=1+a+b-6=0 → a+b=5. P(-2)=-8+4a-2b-6=0 → 4a-2b=14 → 2a-b=7. Solving: a+b=5, 2a-b=7 → 3a=12 → a=4, b=1. Options could be a=4,b=1; a=1,b=4; a=-4,b=-1; a=-1,b=-4. That works. Let's use that: P(x) = x^3 + ax^2 + bx - 6. Factors (x-1) and (x+2). Find a and b. Correct: a=4, b=1. Options: 0: a=4, b=1; 1: a=1, b=4; 2: a=-4, b=-1; 3: a=-1, b=-4. That is fine. I'll adjust the question accordingly.
The equation |2x - 3| = 5 splits into 2x - 3 = 5 or 2x - 3 = -5. Solving gives x = 4 or x = -1.
Using logarithm laws, log_2(16x^3) - log_2(2x) = log_2(16x^3/(2x)) = log_2(8x^2) = log_2(8) + log_2(x^2) = 3 + 2 log_2(x).
The second derivative is y'' = 6x - 12. Setting y'' = 0 gives x = 2. Since y'' changes sign at x = 2, this is a point of inflection.
First find f^{-1}(x): set y = 2x + 1, so x = (y - 1)/2. Thus f^{-1}(x) = (x - 1)/2. Then f^{-1}g(x) = f^{-1}(g(x)) = (g(x) - 1)/2 = (x^2 - 3 - 1)/2 = (x^2 - 4)/2. Set this equal to 1: (x^2 - 4)/2 = 1 → x^2 - 4 = 2 → x^2 = 6 → x = ±√6. This does not match any option. So the question is flawed. Let me adjust: maybe f^{-1}g(x) = 1 with different functions or target. Let's choose f(x) = 3x - 2, g(x) = x^2 + 1. Then f^{-1}(x) = (x + 2)/3. f^{-1}g(x) = (x^2 + 1 + 2)/3 = (x^2 + 3)/3. Set = 1 → x^2 + 3 = 3 → x^2 = 0 → x = 0. Options could be x=0, x=1, x=-1, x=±1. That works. Let's use that: f(x)=3x-2, g(x)=x^2+1. Find x such that f^{-1}g(x)=1. Correct: x=0. Options: 0: x=0; 1: x=1; 2: x=-1; 3: x=±1. That is fine.
Take logarithms of both sides: (2x + 1)log 3 = (x - 2)log 5. Expanding gives 2x log 3 + log 3 = x log 5 - 2 log 5. Rearranging: x(2 log 3 - log 5) = -2 log 5 - log 3. Thus x = (-2 log 5 - log 3)/(2 log 3 - log 5). Using log 3 ≈ 0.4771 and log 5 ≈ 0.6990, numerator ≈ -1.3980 - 0.4771 = -1.8751; denominator ≈ 0.9542 - 0.6990 = 0.2552; x ≈ -7.35? Wait, compute: -1.8751/0.2552 ≈ -7.35, not -3.15. Let me recalc: 2 log 3 = 0.9542, minus log 5 = 0.2552. Numerator: -2 log 5 - log 3 = -1.3980 - 0.4771 = -1.8751. Division gives -7.35. That is not among options. So my equation gives a different answer. Let's solve 3^(2x+1) = 5^(x-2) correctly: (2x+1)ln3 = (x-2)ln5 → 2x ln3 + ln3 = x ln5 - 2 ln5 → x(2ln3 - ln5) = -2ln5 - ln3 → x = (-2ln5 - ln3)/(2ln3 - ln5). Numerically: ln3=1.0986, ln5=1.6094. Numerator: -3.2189 - 1.0986 = -4.3175. Denominator: 2.1972 - 1.6094 = 0.5878. x ≈ -7.345. Not in options. So the question is flawed. I need to choose an equation that yields one of the options. Let's design an equation that gives x ≈ -3.15. Suppose 2^(x+1) = 3^(2x-1). Then (x+1)ln2 = (2x-1)ln3 → x ln2 + ln2 = 2x ln3 - ln3 → x(ln2 - 2ln3) = -ln3 - ln2 → x = (-ln3 - ln2)/(ln2 - 2ln3). Numerically: ln2=0.6931, ln3=1.0986. Numerator: -1.0986 - 0.6931 = -1.7917. Denominator: 0.6931 - 2.1972 = -1.5041. x ≈ 1.19. Not -3.15. Let's try 5^(x+1) = 2^(3x-2). (x+1)ln5 = (3x-2)ln2 → x ln5 + ln5 = 3x ln2 - 2 ln2 → x(ln5 - 3ln2) = -2ln2 - ln5 → x = (-2ln2 - ln5)/(ln5 - 3ln2). ln2=0.6931, ln5=1.6094. Numerator: -1.3863 - 1.6094 = -2.9957. Denominator: 1.6094 - 2.0794 = -0.4700. x ≈ 6.37. Not. Maybe I should change the options to match the correct answer of the original equation. The original equation 3^(2x+1) = 5^(x-2) gives x ≈ -7.35. I can set options to include -7.35. But then need to ensure only one correct. Let's set options: -7.35, -7.53, -7.25, -7.45. But that seems too far. Alternatively, I can change the equation to something simpler: 2^(x+1) = 3^(x-1). Then (x+1)ln2 = (x-1)ln3 → x ln2 + ln2 = x ln3 - ln3 → x(ln2 - ln3) = -ln3 - ln2 → x = (-ln3 - ln2)/(ln2 - ln3). ln3=1.0986, ln2=0.6931. Numerator: -1.0986 - 0.6931 = -1.7917. Denominator: 0.6931 - 1.0986 = -0.4055. x ≈ 4.42. Not. Let's try 3^(x+1) = 5^(x-1). (x+1)ln3 = (x-1)ln5 → x ln3 + ln3 = x ln5 - ln5 → x(ln3 - ln5) = -ln5 - ln3 → x = (-ln5 - ln3)/(ln3 - ln5). ln5=1.6094, ln3=1.0986. Numerator: -1.6094 - 1.0986 = -2.7080. Denominator: 1.0986 - 1.6094 = -0.5108. x ≈ 5.30. Not. I think it's best to replace this question with a different one that is not an exponential equation requiring logs, or use a simpler exponential equation that yields a nice answer. For example, solve 2^(x+1) = 8^(x-1). Since 8 = 2^3, we have 2^(x+1) = 2^(3x-3), so x+1 = 3x-3 → 2x = 4 → x = 2. That is easy. But we already have an exponential equation. Maybe use a different topic: solve log_2(x+3) + log_2(x-1) = 3. Then log_2((x+3)(x-1)) = 3 → (x+3)(x-1) = 8 → x^2 + 2x - 3 = 8 → x^2 + 2x - 11 = 0 → x = [-2 ± √(4+44)]/2 = [-2 ± √48]/2 = -1 ± 2√3. Since x must be >1, x = -1 + 2√3 ≈ 2.46. Not nice. Maybe log_2(x+1) + log_2(x-1) = 3 → (x+1)(x-1)=8 → x^2 -1=8 → x^2=9 → x=3 (since x>1). That works. So question: Solve log_2(x+1) + log_2(x-1) = 3. Options: x=3, x=-3, x=3 or x=-3, x=9. Correct: x=3. That is a good medium question. I'll replace the exponential equation question with this logarithm equation question. But I already have a logarithm laws question. This is a logarithmic equation, which is a different sub-topic. That's fine. So I will use: Solve log_2(x+1) + log_2(x-1) = 3. Options: 0: x=3; 1: x=-3; 2: x=3 or x=-3; 3: x=9. CorrectIndex 0. Explanation: Combine logs: log_2((x+1)(x-1))=3 → (x+1)(x-1)=8 → x^2-1=8 → x^2=9 → x=±3. But x must be >1 for both logs to be defined, so x=3. Wrong: 1: x=-3 makes the arguments negative, so logs undefined. 2: includes the invalid solution. 3: x=9 gives log_2(10)+log_2(8)=log_2(80)≠3. So that works.
Complete the square: f(x) = (x - 3)^2 - 4. Since x ≥ 3, (x - 3)^2 ≥ 0, so f(x) ≥ -4. The minimum value is -4 at x = 3.
Factorise: (x - 2)(x - 3) > 0. The critical values are x = 2 and x = 3. Testing intervals shows the product is positive when x < 2 or x > 3.
First find f^{-1}(x): y = 2x + 3 → x = (y - 3)/2, so f^{-1}(x) = (x - 3)/2. Set equal to g(x): (x - 3)/2 = x^2 - 1. Multiply by 2: x - 3 = 2x^2 - 2 → 2x^2 - x + 1 = 0? Wait, rearrange: x - 3 = 2x^2 - 2 → 2x^2 - x + 1 = 0. Discriminant = 1 - 8 = -7 < 0, no real solutions. So this is flawed. Let me adjust: maybe f(x) = 2x - 3. Then f^{-1}(x) = (x + 3)/2. Set = x^2 - 1: (x + 3)/2 = x^2 - 1 → x + 3 = 2x^2 - 2 → 2x^2 - x - 5 = 0. Discriminant 1 + 40 = 41, not nice. Let's choose f(x) = 2x + 1, g(x) = x^2 - 1. f^{-1}(x) = (x - 1)/2. Set = x^2 - 1: (x - 1)/2 = x^2 - 1 → x - 1 = 2x^2 - 2 → 2x^2 - x - 1 = 0 → (2x + 1)(x - 1) = 0 → x = -1/2 or x = 1. Not integer. Maybe f(x) = 3x - 2, g(x) = x^2. f^{-1}(x) = (x + 2)/3. Set = x^2: (x + 2)/3 = x^2 → x + 2 = 3x^2 → 3x^2 - x - 2 = 0 → (3x + 2)(x - 1) = 0 → x = -2/3 or x = 1. Not nice. Let's try f(x) = x + 2, g(x) = x^2 - 4. f^{-1}(x) = x - 2. Set = x^2 - 4: x - 2 = x^2 - 4 → x^2 - x - 2 = 0 → (x - 2)(x + 1) = 0 → x = 2 or x = -1. That works. So question: f(x) = x + 2, g(x) = x^2 - 4. Find x such that f^{-1}(x) = g(x). Options: 0: x = 2 or x = -1; 1: x = -2 or x = 1; 2: x = 2 only; 3: x = -1 only. CorrectIndex 0. Explanation: f^{-1}(x) = x - 2. Set x - 2 = x^2 - 4 → x^2 - x - 2 = 0 → (x - 2)(x + 1) = 0 → x = 2 or x = -1. Wrong: 1: signs reversed; 2: omits -1; 3: omits 2. Good.
Divide P(x) by (x - 2) to get x^2 + 4x + 3. Factorising gives (x + 1)(x + 3). So P(x) = (x - 2)(x + 1)(x + 3).
Set x^2 - 2x + 3 = 2x + 1, giving x^2 - 4x + 2 = 0. Solving: x = [4 ± √(16 - 8)]/2 = [4 ± √8]/2 = 2 ± √2. This does not give integer coordinates. So the question is flawed. Let me choose a different line and curve that yield integer intersections. For example, line y = x + 1 and curve y = x^2 - 1. Set x^2 - 1 = x + 1 → x^2 - x - 2 = 0 → (x - 2)(x + 1) = 0 → x = 2 or x = -1. Then y = 3 or y = 0. Points: (2, 3) and (-1, 0). Options: 0: (2, 3) and (-1, 0); 1: (2, 3) only; 2: (-1, 0) only; 3: (2, 3) and (1, 2). That works. So question: Find the coordinates of the points of intersection of the line y = x + 1 and the curve y = x^2 - 1. Correct: (2, 3) and (-1, 0).
To find the inverse, set y = 2x - 5 and solve for x: add 5 to both sides to get y + 5 = 2x, then divide by 2 to obtain x = (y + 5)/2. Finally, swap x and y to write f^{-1}(x) = (x + 5)/2. This is the only expression that undoes the operations of f.
By the Remainder Theorem, the remainder is P(3). Compute P(3) = 3^3 - 4(3^2) + 5(3) - 2 = 27 - 36 + 15 - 2 = 4. Thus the remainder is 4.
Express both sides with base 2: 8 = 2^3, so 8^(x-1) = (2^3)^(x-1) = 2^(3x-3). The equation becomes 2^(x+1) = 2^(3x-3). Since the bases are equal, equate exponents: x + 1 = 3x - 3, which gives 4 = 2x, so x = 2.
Use the logarithm product rule: log_3 27 + log_3 9 = log_3(27 × 9) = log_3 243. Since 3^5 = 243, log_3 243 = 5. Alternatively, log_3 27 = 3 and log_3 9 = 2, so the sum is 5.
Complete the square: x^2 + 6x + 5 = (x^2 + 6x + 9) - 9 + 5 = (x + 3)^2 - 4. Thus a = 3 and b = -4.
Complete the square: y = (x - 2)^2 + 3. The minimum value occurs when the squared term is zero, i.e., x = 2, giving y = 3. So the minimum point is (2, 3). Alternatively, use x = -b/(2a) = 4/2 = 2 and substitute to find y = 3.
Factorise: x^2 - 9 = (x - 3)(x + 3). The critical values are x = -3 and x = 3. Since the coefficient of x^2 is positive, the parabola opens upwards, so the expression is negative between the roots and zero at the roots. Hence -3 ≤ x ≤ 3.
First evaluate g(3) = 3^2 = 9. Then apply f to this result: f(9) = 2(9) + 1 = 19. So f(g(3)) = 19.
Write 16 as a power of 2: 16 = 2^4. Then the equation becomes 2^(x+1) = 2^4. Equating exponents gives x + 1 = 4, so x = 3.
The equation |2x - 3| = 7 splits into two cases: 2x - 3 = 7 or 2x - 3 = -7. Solving the first gives 2x = 10 so x = 5. Solving the second gives 2x = -4 so x = -2. Both solutions satisfy the original equation.
Find the first derivative: dy/dx = 3x^2 - 6x = 3x(x - 2). Stationary points occur at x = 0 and x = 2. The second derivative is d^2y/dx^2 = 6x - 6. At x = 0, the second derivative is -6, which is negative, indicating a local maximum.
Let y = 3^x. Then the equation becomes y^2 - 10y + 9 = 0, which factorises as (y - 1)(y - 9) = 0, so y = 1 or y = 9. Since y = 3^x, we have 3^x = 1 giving x = 0, and 3^x = 9 giving x = 2.
Complete the square: f(x) = (x - 2)^2 + 3. Since x ≥ 2, the term (x - 2)^2 is always non-negative, and its minimum value is 0 when x = 2. Therefore the minimum value of f is 3, and the range is f(x) ≥ 3.
For two distinct real roots, the discriminant must be positive: Δ = (k - 3)^2 - 4(1)(k) > 0. Expanding gives k^2 - 6k + 9 - 4k = k^2 - 10k + 9 > 0. Factorising: (k - 1)(k - 9) > 0. This inequality holds when k < 1 or k > 9.
First find f^{-1}(x): set y = 2x + 3, then x = (y - 3)/2, so f^{-1}(x) = (x - 3)/2. Set this equal to g(x): (x - 3)/2 = x^2 - 1. Multiply by 2: x - 3 = 2x^2 - 2, which rearranges to 2x^2 - x + 1 = 0. Wait, that gives discriminant (-1)^2 - 4(2)(1) = 1 - 8 = -7 < 0, so no real solutions. Let me recalc: (x - 3)/2 = x^2 - 1 → x - 3 = 2x^2 - 2 → 2x^2 - x + 1 = 0. Indeed no real roots. I need to check the question. Perhaps I intended f^{-1}(x) = g(x) but with different functions. Let me choose f(x) = 2x - 1 and g(x) = x^2 - 4. Then f^{-1}(x) = (x+1)/2. Setting equal: (x+1)/2 = x^2 - 4 → x+1 = 2x^2 - 8 → 2x^2 - x - 9 = 0. Discriminant 1+72=73, not nice. I need a clean quadratic. Let me choose f(x) = 2x + 1 and g(x) = x^2 - 2. Then f^{-1}(x) = (x-1)/2. Set (x-1)/2 = x^2 - 2 → x-1 = 2x^2 - 4 → 2x^2 - x - 3 = 0 → (2x - 3)(x + 1) = 0 → x = 3/2 or x = -1. That works. So I'll change the question to f(x)=2x+1, g(x)=x^2-2. Then the correct answer is x = 3/2 or x = -1. But the options need to match. Let me adjust.
The critical values are where the numerator is zero (x = -1) and where the denominator is zero (x = 3). Test intervals: for x < -1, both numerator and denominator are negative, so the quotient is positive. For -1 < x < 3, the numerator is positive and the denominator negative, so the quotient is negative. For x > 3, both are positive, so the quotient is positive. At x = -1 the quotient is zero, which satisfies ≥ 0. At x = 3 the expression is undefined. Hence the solution is x ≤ -1 or x > 3.
Use the laws of logarithms: log_a (x^3 y^2 / z) = log_a (x^3) + log_a (y^2) - log_a z. Then apply the power rule: log_a (x^3) = 3 log_a x and log_a (y^2) = 2 log_a y. Thus the expression becomes 3 log_a x + 2 log_a y - log_a z.
Complete the square: f(x) = (x + 1)^2 + 4. Let y = (x + 1)^2 + 4. Then (x + 1)^2 = y - 4, so x + 1 = ±√(y - 4). Since the domain is x ≥ -1, we have x + 1 ≥ 0, so we take the positive root: x + 1 = √(y - 4). Thus x = -1 + √(y - 4), and swapping x and y gives f^{-1}(x) = -1 + √(x - 4).
Find the second derivative: y' = 3x^2 - 12x + 9, y'' = 6x - 12. Set y'' = 0: 6x - 12 = 0 ⇒ x = 2. Since y'' changes sign at x = 2 (negative for x < 2, positive for x > 2), this is a point of inflection.
Let y = 2^x. Then the equation becomes y^2 - 5y + 4 = 0, which factorises as (y - 1)(y - 4) = 0, so y = 1 or y = 4. Since y = 2^x, we have 2^x = 1 giving x = 0, and 2^x = 4 giving x = 2.
Find the first derivative: dy/dx = 3x^2 - 6x = 3x(x - 2). At x = 0, dy/dx = 0, so it is a stationary point. The second derivative is d^2y/dx^2 = 6x - 6. At x = 0, this is -6, which is negative, indicating a local maximum.
To find the inverse, start with y = 3x - 2 and solve for x: add 2 to both sides to get y + 2 = 3x, then divide by 3, giving x = (y + 2)/3. Finally, interchange x and y to write the inverse function as f^{-1}(x) = (x + 2)/3. This is the correct inverse because f((x+2)/3) = 3((x+2)/3) - 2 = x.
First compute fg(x) = f(g(x)) = f(x + 1) = (x + 1)^2. Then compute gf(x) = g(f(x)) = g(x^2) = x^2 + 1. Setting these equal gives (x + 1)^2 = x^2 + 1, which expands to x^2 + 2x + 1 = x^2 + 1. Subtracting x^2 + 1 from both sides leaves 2x = 0, so x = 0. Checking: fg(0) = f(1) = 1 and gf(0) = g(0) = 1, so the solution is confirmed.
Using the product law for logarithms, log_2(x) + log_2(x - 2) = log_2[x(x - 2)]. The equation becomes log_2[x(x - 2)] = 3. Converting to exponential form gives x(x - 2) = 2^3 = 8, so x^2 - 2x - 8 = 0. Factorising gives (x - 4)(x + 2) = 0, so x = 4 or x = -2. However, logarithms are only defined for positive arguments, so x must satisfy x > 0 and x - 2 > 0, i.e. x > 2. Thus x = -2 is rejected, leaving x = 4 as the only solution.
The notation fg(x) means f applied to the output of g, so substitute g(x) into f: f(x^2 + 2x) = 5(x^2 + 2x) - 4 = 5x^2 + 10x - 4. The whole expression g(x) replaces x in f, and the constant -4 is applied only once, after the multiplication by 5.