Calculus Midterm Prep

5. Monomial Exponent Rules: \(\frac{(3x^3 y^{-2})^3}{9x^4 y^{-8}}\) Key: A

Cube numerator: \(3^3 (x^3)^3 (y^{-2})^3 = 27 x^9 y^{-6}\). Divide by \(9x^4 y^{-8}\):
\(\frac{27}{9} \cdot x^{9 - 4} \cdot y^{-6 - (-8)} = 3 x^5 y^2\).

6. Monomial Exponent Rules: \(\left(\frac{2a^{-2} b^4}{6a^3 b^{-1}}\right)^{-2}\) Key: A

Simplify inside parentheses: \(\frac{2}{6} \cdot a^{-2 - 3} \cdot b^{4 - (-1)} = \frac{1}{3} a^{-5} b^5 = \frac{b^5}{3a^5}\). Invert and square: \(\left(\frac{3a^5}{b^5}\right)^2 = \frac{9a^{10}}{b^{10}}\).

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7. Domain of \(f(x) = \frac{\sqrt{6 - x}}{2x + 10}\) Key: A

1. Radicand condition: \(6 - x \ge 0 \implies x \le 6\).
2. Denominator condition: \(2x + 10 \ne 0 \implies x \ne -5\).
   Intersection: \((-\infty, -5) \cup (-5, 6]\).

8. Domain of \(g(x) = \frac{\sqrt{3x + 15}}{x^2 - 25}\) Key: B

1. Radicand: \(3x + 15 \ge 0 \implies x \ge -5\).
2. Denominator: \(x^2 - 25 \ne 0 \implies x \ne 5\) and \(x \ne -5\).
   Since \(x = -5\) makes the denominator zero, \(-5\) is excluded: \((-5, 5) \cup (5, \infty)\).

9. Synthetic Division: \(P(x) = 2x^3 - 3x^2 - 8x - 3\) given \((x - 3)\) Key: A

Divide coefficients \([2, -3, -8, -3]\) by \(3\): yields quotient \(2x^2 + 3x + 1\).
Factor quadratic: \((2x + 1)(x + 1)\). Complete factorization: \((x - 3)(2x + 1)(x + 1)\).

10. Synthetic Division: \(Q(x) = 3x^3 - 4x^2 - 17x + 6\) given \((x + 2)\) Key: C

Divide coefficients \([3, -4, -17, 6]\) by \(-2\): yields quotient \(3x^2 - 10x + 3\).
Factor quadratic: \((3x - 1)(x - 3)\). Complete factorization: \((x + 2)(3x - 1)(x - 3)\).

11. Difference of Cubes: \(64x^3 - 27y^3\) Key: A

Use identity \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) with \(a = 4x\) and \(b = 3y\):
\((4x - 3y)((4x)^2 + (4x)(3y) + (3y)^2) = (4x - 3y)(16x^2 + 12xy + 9y^2)\).

12. Factoring by Grouping: \(2x^3 - 3x^2 - 18x + 27\) Key: A

Group terms: \(x^2(2x - 3) - 9(2x - 3) = (2x - 3)(x^2 - 9)\).
Factor difference of squares: \((2x - 3)(x + 3)(x - 3)\).

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13. Fractional Algebraic Powers: \(\frac{-3x}{(x + 2)^{3/2}} - \frac{2}{(x + 2)^{1/2}}\) Key: A

Common denominator \((x + 2)^{3/2}\): multiply second term by \(\frac{x + 2}{x + 2}\):
\(\frac{-3x - 2(x + 2)}{(x + 2)^{3/2}} = \frac{-3x - 2x - 4}{(x + 2)^{3/2}} = \frac{-5x - 4}{(x + 2)^{3/2}}\).

14. Fractional Algebraic Powers: \(\frac{5x}{(2x - 1)^{3/2}} - \frac{3}{(2x - 1)^{1/2}}\) Key: B

Common denominator \((2x - 1)^{3/2}\):
\(\frac{5x - 3(2x - 1)}{(2x - 1)^{3/2}} = \frac{5x - 6x + 3}{(2x - 1)^{3/2}} = \frac{-x + 3}{(2x - 1)^{3/2}}\).

15. Break-Even Business Model 1 Key: B

Fixed costs \(F = \$31,500\); selling price \(P = \$24.70\); unit variable cost \(V = \$14.20\).
Unit contribution margin \(= P - V = 24.70 - 14.20 = \$10.50\).
Break-even volume: \(x = \frac{31,500}{10.50} = 3,000\text{ units}\).

16. Break-Even Business Model 2 Key: C

Fixed equipment costs \(F = \$22,400\); selling price \(P = \$7.00\); variable cost \(V = \$3.80\).
Contribution margin per loaf \(= 7.00 - 3.80 = \$3.20\).
Break-even volume: \(x = \frac{22,400}{3.20} = 7,000\text{ loaves}\).

17. Rational Root Theorem: \(x^3 - 19x - 30\) Key: A

Factors of \(30\): \(\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30\).
Test \(x = -2\): \((-2)^3 - 19(-2) - 30 = -8 + 38 - 30 = 0\). Quotient is \(x^2 - 2x - 15\).
Factoring quotient: \((x - 5)(x + 3) = 0 \implies x = 5, -3\). Complete real zeros: \(-3, -2, 5\).

18. Rational Root Theorem: \(x^3 - 13x + 12\) Key: A

Test \(x = 1\): \(1^3 - 13(1) + 12 = 0\). Divide by \((x - 1)\) gives \(x^2 + x - 12\).
Factoring quotient: \((x + 4)(x - 3) = 0 \implies x = -4, 3\). Real zeros: \(-4, 1, 3\).

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19. Circle Equation: \(x^2 + y^2 - 6x + 10y + 9 = 0\) Key: A

Complete the square: \((x^2 - 6x + 9) + (y^2 + 10y + 25) = -9 + 9 + 25 = 25\).
Standard form: \((x - 3)^2 + (y + 5)^2 = 5^2\). Center is \((3, -5)\) and radius is \(r = 5\).

20. Circle with Coefficients: \(4x^2 + 4y^2 + 32x - 16y - 20 = 0\) Key: C

Divide entire equation by 4: \(x^2 + y^2 + 8x - 4y - 5 = 0\).
Complete square: \((x^2 + 8x + 16) + (y^2 - 4y + 4) = 5 + 16 + 4 = 25\).
Standard form: \((x + 4)^2 + (y - 2)^2 = 5^2\). Center: \((-4, 2)\), Radius: \(r = 5\).

21. Difference Quotient: \(f(x) = 4x^2 - 5x + 3\) Key: A

\(f(x+h) - f(x) = [4(x^2 + 2xh + h^2) - 5(x+h) + 3] - [4x^2 - 5x + 3] = 8xh + 4h^2 - 5h\).
Dividing by \(h\) yields \(\frac{h(8x + 4h - 5)}{h} = 8x + 4h - 5\).

22. Difference Quotient: \(g(x) = -3x^2 + 8x - 2\) Key: B

\(g(x+h) - g(x) = [-3(x^2 + 2xh + h^2) + 8x + 8h - 2] - [-3x^2 + 8x - 2] = -6xh - 3h^2 + 8h\).
Dividing by \(h\) yields \(-6x - 3h + 8\).

23. Polynomial Inverse: \(f(x) = 2x^3 - 7\) Key: A

Set \(y = 2x^3 - 7\) and isolate \(x\): \(2x^3 = y + 7 \implies x^3 = \frac{y + 7}{2} \implies x = \sqrt[3]{\frac{y + 7}{2}}\).
In terms of \(x\): \(f^{-1}(x) = \sqrt[3]{\frac{x + 7}{2}}\).

24. Polynomial Inverse: \(g(x) = 4x^3 + 5\) Key: C

Set \(y = 4x^3 + 5\) and isolate \(x\): \(4x^3 = y - 5 \implies x^3 = \frac{y - 5}{4} \implies x = \sqrt[3]{\frac{y - 5}{4}}\).
In terms of \(x\): \(g^{-1}(x) = \sqrt[3]{\frac{x - 5}{4}}\).

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25. Factoring Limit: \(\lim_{x \to 4} \frac{x^2 - x - 12}{x^2 - 16}\) Key: A

Indeterminate form \(\frac{0}{0}\). Factor: \(\lim_{x \to 4} \frac{(x - 4)(x + 3)}{(x - 4)(x + 4)} = \lim_{x \to 4} \frac{x + 3}{x + 4}\).
Direct substitution yields: \(\frac{4 + 3}{4 + 4} = \frac{7}{8}\).

26. Direct Substitution Limit: \(\lim_{x \to 2} \frac{3x^2 - 4x + 1}{2x + 3}\) Key: A

Denominator at \(x = 2\) is \(2(2) + 3 = 7 \ne 0\). Evaluate directly by substitution:
\(\frac{3(2)^2 - 4(2) + 1}{2(2) + 3} = \frac{3(4) - 8 + 1}{4 + 3} = \frac{12 - 8 + 1}{7} = \frac{5}{7}\).

27. Radical Conjugate Limit: \(\lim_{x \to 7} \frac{\sqrt{x + 9} - 4}{x - 7}\) Key: A

Multiply by \((\sqrt{x + 9} + 4)\): \(\lim_{x \to 7} \frac{(x + 9) - 16}{(x - 7)(\sqrt{x + 9} + 4)} = \lim_{x \to 7} \frac{x - 7}{(x - 7)(\sqrt{x + 9} + 4)}\).
Cancel \((x - 7)\): \(\frac{1}{\sqrt{7 + 9} + 4} = \frac{1}{4 + 4} = \frac{1}{8}\).

28. Radical Conjugate Limit: \(\lim_{x \to 0} \frac{\sqrt{2x + 25} - 5}{x}\) Key: C

Multiply by \((\sqrt{2x + 25} + 5)\): \(\lim_{x \to 0} \frac{2x + 25 - 25}{x(\sqrt{2x + 25} + 5)} = \lim_{x \to 0} \frac{2x}{x(\sqrt{2x + 25} + 5)}\).
Cancel \(x\): \(\frac{2}{\sqrt{25} + 5} = \frac{2}{10} = \frac{1}{5}\).

29. Continuity: \(g(x) = \frac{x^2 - 9}{x^2 - 7x + 12}\) Key: A

Denominator: \((x - 3)(x - 4) = 0 \implies x = 3, 4\). Discontinuities at \(x = 3\) (removable hole) and \(x = 4\) (vertical asymptote).
Continuous on its domain: \((-\infty, 3) \cup (3, 4) \cup (4, \infty)\).

30. Continuity: \(h(x) = \frac{2x - 3}{x^2 - 5x}\) Key: B

Denominator: \(x(x - 5) = 0 \implies x = 0, 5\). Discontinuous at \(0\) and \(5\).
Continuous everywhere else on the real line: \((-\infty, 0) \cup (0, 5) \cup (5, \infty)\).

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31. Product & Quotient Rule: \(f(x) = \frac{(2x + 1)(x - 3)}{x + 2}\) Key: A

Expand numerator: \(u(x) = 2x^2 - 5x - 3 \implies u'(x) = 4x - 5\). Denominator: \(v(x) = x + 2 \implies v'(x) = 1\).
Quotient rule: \(\frac{u'v - uv'}{v^2} = \frac{(4x - 5)(x + 2) - (2x^2 - 5x - 3)(1)}{(x + 2)^2}\)
\(= \frac{(4x^2 + 3x - 10) - (2x^2 - 5x - 3)}{(x + 2)^2} = \frac{2x^2 + 8x - 7}{(x + 2)^2}\).

32. Find \(f'(x)\) for: \(f(x) = (3x^2 - 2)(5x + 3)\) Key: A

Product rule \(u'v + uv'\): \(f'(x) = (6x)(5x + 3) + (3x^2 - 2)(5) = 30x^2 + 18x + 15x^2 - 10 = 45x^2 + 18x - 10\).

33. Quotient Rule: \(f(x) = \frac{4x - 3}{2x + 5}\) Key: A

Quotient rule \(\frac{u'v - uv'}{v^2}\): \(\frac{4(2x + 5) - (4x - 3)(2)}{(2x + 5)^2} = \frac{8x + 20 - 8x + 6}{(2x + 5)^2} = \frac{26}{(2x + 5)^2}\).

34. General Power Rule: \(g(x) = (2x^2 - 7x + 4)^5\) Key: A

Chain rule: \(5u^4 \cdot u'\) where \(u = 2x^2 - 7x + 4 \implies u' = 4x - 7\).
\(g'(x) = 5(4x - 7)(2x^2 - 7x + 4)^4\).

35. Radical Times Binomial Cubed: \(f(x) = \sqrt{x}(2x + 1)^3\) Key: A

Product rule + chain rule: \(f'(x) = \frac{1}{2\sqrt{x}}(2x + 1)^3 + \sqrt{x} \cdot 3(2x + 1)^2(2)\)
\(= \frac{(2x + 1)^3}{2\sqrt{x}} + 6\sqrt{x}(2x + 1)^2 = \frac{(2x + 1)^2 [(2x + 1) + 12x]}{2\sqrt{x}} = \frac{(2x + 1)^2(14x + 1)}{2\sqrt{x}}\).

36. Derivative: \(k(x) = \frac{1}{\sqrt{3x^2 + 7}}\) Key: C

\(k(x) = (3x^2 + 7)^{-1/2} \implies k'(x) = -\frac{1}{2}(3x^2 + 7)^{-3/2}(6x) = \frac{-3x}{(3x^2 + 7)^{3/2}}\).

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Section II

Open Response Solutions: Distance & Systems

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37. Distance: \((x, 5)\) & \((-2, 9)\) with \(d = 5\) Answer: \(x = 1\) or \(x = -5\)

Formula: \(\sqrt{(x - (-2))^2 + (9 - 5)^2} = 5 \implies (x + 2)^2 + 16 = 25 \implies (x + 2)^2 = 9\).
Square root: \(x + 2 = \pm 3 \implies x = -2 \pm 3 \implies x = 1\) or \(x = -5\).

Rubric: +1 pt: Setup distance eq. +1 pt: Simplify to \((x+2)^2 = 9\) +1 pt: Both roots \(x = 1, -5\)

38. Distance: \((3, y)\) & \((-5, 2)\) with \(d = 10\) Answer: \(y = 8\) or \(y = -4\)

Formula: \(\sqrt{(3 - (-5))^2 + (y - 2)^2} = 10 \implies 8^2 + (y - 2)^2 = 100 \implies 64 + (y - 2)^2 = 100\).
\((y - 2)^2 = 36 \implies y - 2 = \pm 6 \implies y = 2 \pm 6 \implies y = 8\) or \(y = -4\).

Rubric: +1 pt: Setup distance eq. +1 pt: Simplify to \((y-2)^2 = 36\) +1 pt: Both roots \(y = 8, -4\)

39. System: \(y = x^2 - 5x + 4\) & \(y = 2x - 6\) Answer: \((2, -2)\) and \((5, 4)\)

Equate: \(x^2 - 5x + 4 = 2x - 6 \implies x^2 - 7x + 10 = 0 \implies (x - 2)(x - 5) = 0 \implies x = 2, 5\).
Substitute into \(y = 2x - 6\): for \(x = 2\), \(y = -2 \implies (2, -2)\); for \(x = 5\), \(y = 4 \implies (5, 4)\).

Rubric: +1 pt: Standard form \(x^2 - 7x + 10 = 0\) +1 pt: Solve \(x = 2, 5\) +1 pt: Points \((2, -2)\) and \((5, 4)\)

40. System: \(y = 2x^2 - 4x - 1\) & \(y = 3x + 3\) Answer: \((4, 15)\) and \((-1/2, 3/2)\)

Equate: \(2x^2 - 4x - 1 = 3x + 3 \implies 2x^2 - 7x - 4 = 0 \implies (2x + 1)(x - 4) = 0 \implies x = 4, -1/2\).
\(y = 3x + 3\): for \(x = 4\), \(y = 15 \implies (4, 15)\); for \(x = -1/2\), \(y = 3/2 \implies (-1/2, 3/2)\).

Rubric: +1 pt: Standard form \(2x^2 - 7x - 4 = 0\) +1 pt: Solve \(x = 4, -1/2\) +1 pt: Points \((4, 15)\) and \((-1/2, 3/2)\)

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Section II

Open Response Solutions: Tangents & Extrema

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41. Tangent Line: \(f(x) = 2x^3 - 4x^2 + 3\) at \((-1, -3)\) Answer: \(y = 14x + 11\)

1. Derivative: \(f'(x) = 6x^2 - 8x\).
2. Slope at \(x = -1\): \(m = f'(-1) = 6(-1)^2 - 8(-1) = 6 + 8 = 14\).
3. Point-slope: \(y - (-3) = 14(x - (-1)) \implies y + 3 = 14x + 14 \implies y = 14x + 11\).

Rubric: +1 pt: Derivative \(6x^2 - 8x\) +1 pt: Slope \(m = 14\) +1 pt: Line \(y = 14x + 11\)

42. Tangent Line: \(g(x) = x^3 - 6x + 2\) at \((2, -2)\) Answer: \(y = 6x - 14\)

1. Derivative: \(g'(x) = 3x^2 - 6\).
2. Slope at \(x = 2\): \(m = g'(2) = 3(4) - 6 = 6\).
3. Point-slope: \(y - (-2) = 6(x - 2) \implies y + 2 = 6x - 12 \implies y = 6x - 14\).

Rubric: +1 pt: Derivative \(3x^2 - 6\) +1 pt: Slope \(m = 6\) +1 pt: Line \(y = 6x - 14\)

43. Maximum Height: \(s(t) = -4.9t^2 + 44.1t + 80\) Answer: \(t = 4.5\text{ seconds}\)

1. Velocity: \(v(t) = s'(t) = -9.8t + 44.1\).
2. Max height condition: \(v(t) = 0 \implies 9.8t = 44.1 \implies t = \frac{44.1}{9.8} = 4.5\text{ seconds}\).
3. \(s''(t) = -9.8 < 0\), confirming maximum by second derivative test.

Rubric: +1 pt: Velocity \(v(t) = -9.8t + 44.1\) +1 pt: Setting \(v = 0\) & solving +1 pt: Correct time \(t = 4.5\text{ s}\)

44. Maximum Altitude: \(h(t) = -4.9t^2 + 49.0t\) Answer: \(t = 5.0\text{ s}, h_{\text{max}} = 122.5\text{ m}\)

1. Velocity: \(v(t) = h'(t) = -9.8t + 49.0\).
2. Critical value: \(v(t) = 0 \implies 9.8t = 49.0 \implies t = \frac{49.0}{9.8} = 5.0\text{ seconds}\).
3. Maximum altitude: \(h(5) = -4.9(25) + 49.0(5) = -122.5 + 245 = 122.5\text{ meters}\).

Rubric: +1 pt: Velocity \(v(t) = -9.8t + 49.0\) +1 pt: Time \(t = 5.0\text{ s}\) +1 pt: Max altitude \(122.5\text{ m}\)

Applied Calculus Mid-Semester Exam Review Solutions End of Review Packet Solutions Key