The correct answer is D (14). Substitute x = −2 into the function: f(−2) = (−2)² − 3(−2) + 4 = 4 + 6 + 4 = 14. The three key steps are: (1) square the input: (−2)² = +4, not −4; (2) multiply: −3(−2) = +6, not −6; (3) add: 4 + 6 + 4 = 14. A (2) comes from treating −3(−2) as −6: 4 − 6 + 4 = 2. B (6) comes from treating (−2)² as −4: −4 + 6 + 4 = 6. C (10) comes from a partial sign error. Pro tip: when substituting a negative value, write every step explicitly — sign errors on squared terms and products are the most common mistakes in function evaluation.

This is a recursive sequences question testing careful step-by-step application of a rule. Choice C (33) is correct — apply the rule aₙ = 2aₙ₋₁ − 1 repeatedly: a₁ = 3, a₂ = 2(3) − 1 = 5, a₃ = 2(5) − 1 = 9, a₄ = 2(9) − 1 = 17, a₅ = 2(17) − 1 = 33. Choice A (17) stops at a₄ — off by one term, likely from miscounting or starting the index at n = 0. Choice B (23) results from an arithmetic error at one step — perhaps computing a₃ = 2(5) + 1 = 11 (adding instead of subtracting): then a₄ = 2(11) − 1 = 21, a₅ = 2(21) − 1 = 41... or some other mid-sequence error. Choice D (65) comes from using aₙ = 2aₙ₋₁ + 1 throughout (adding instead of subtracting): a₂ = 7, a₃ = 15, a₄ = 31, a₅ = 63 ≈ 65. Pro tip: For recursive sequences, build a table of values term by term — there's no shortcut to jumping ahead. Write out each term explicitly: write the term number, then compute carefully. The −1 in this rule is easy to misread as +1, which produces very different values.

This is a piecewise functions question testing domain-aware equation solving. Choice B (1) is correct — solve f(x) = 5 in each piece separately, then check validity. Case 1 (x < 0): 2x + 1 = 5 → x = 2. Invalid: 2 ≥ 0, so it falls outside this piece's domain. Case 2 (x ≥ 0): x² − 3 = 5 → x² = 8 → x = ±2√2 ≈ ±2.83. Check: 2√2 ≥ 0 ✓ (valid); −2√2 < 0 ✗ (invalid for this piece). Exactly 1 valid solution. Choice A (0) results from making errors in both cases and concluding no solution exists. Choice C (2) counts both ±2√2 from Case 2 without checking that −2√2 fails the x ≥ 0 domain requirement. Choice D (3) counts the invalid solution x = 2 from Case 1 plus both ±2√2 from Case 2, accepting all algebraic results without domain verification. Pro tip: For piecewise functions, always solve each piece independently, then check whether each solution satisfies that piece's domain condition. An algebraically valid answer that violates the domain restriction must be rejected. This is the piecewise equivalent of checking for extraneous solutions in radical equations.

We need to find $f(1-x)$ where $f(x) = 2 - x$. This means replacing every x with (1-x): $f(1-x) = 2 - (1-x) = 2 - 1 + x = 1 + x$. When distributing the negative sign, $-(1-x)$ becomes $-1 + x$.

We need to find f(x + 4) when f(x) = 2x² - x. To find f(x + 4), we replace every x in the original function with (x + 4): f(x + 4) = 2(x + 4)² - (x + 4). This expression shows the direct substitution before any expansion. Choice D correctly represents this substitution form. Note that f(x + 4) means substituting (x + 4) for every x, not adding 4 to the result.

We need to find f(-3) when $f(x) = 5x^2 - 4x + 1$. To evaluate f(-3), we substitute x = -3 into the function: $f(-3) = 5(-3)^2 - 4(-3) + 1$. Following order of operations, we calculate $5(-3)^2 = 5(9) = 45$, then $4(-3) = -12$, so $f(-3) = 45 - (-12) + 1 = 45 + 12 + 1 = 58$. Therefore, $f(-3) = 58$.

We need to find f(3) when f(x) = 7 - x. To evaluate f(3), we substitute x = 3 into the function: f(3) = 7 - 3. Following order of operations, we calculate 7 - 3 = 4. Therefore, f(3) = 4.

We need to find f(7) when $f(x) = \frac{3x - 1}{2}$. To evaluate f(7), we substitute x = 7 into the function: $f(7) = \frac{3(7) - 1}{2}$. Following order of operations, we calculate the numerator: $3(7) - 1 = 21 - 1 = 20$, then $20/2 = 10$. Therefore, $f(7) = 10$.

We need to find f(-1) where f(x) = x² + 2x - 5. Substituting x = -1 with parentheses: f(-1) = (-1)² + 2(-1) - 5 = 1 - 2 - 5 = -6. Remember that (-1)² = 1, not -1, since we're squaring the entire negative number.

We need to find f(x) + 1 when f(x) = x² - 1. Since f(x) = x² - 1, we have f(x) + 1 = (x² - 1) + 1 = x² - 1 + 1 = x². The answer is x².