Practice 1

MA150 Final Exam Practice Notes

\n## General Instructions

• Complete the worksheet in your small group.

• Submit the completed worksheet as instructed.

• Show all work and justify answers for full credit.

• Write explanations using complete sentences in the provided lines.
  \n## Limit Evaluations

Problem 1: Evaluate each limit

a) Limit as x approaches 3:

• Expression: $\lim_{x \to 3} \frac{x^3 - 27}{x^2 - 9}$

• Justification: Factor both the numerator and denominator.

  • Numerator: $x^3 - 27 = (x - 3)(x^2 + 3x + 9)$

  • Denominator: $x^2 - 9 = (x - 3)(x + 3)$

• Simplified Limit:

  • $\lim<em>{x \to 3} \frac{(x - 3)(x^2 + 3x + 9)}{(x - 3)(x + 3)} = \lim</em>{x \to 3} \frac{x^2 + 3x + 9}{x + 3}$

• Evaluation: Substitute $x = 3$ into the remaining equation.

  • Result: $\frac{3^2 + 3(3) + 9}{3 + 3} = \frac{9 + 9 + 9}{6} = \frac{27}{6} = \frac{9}{2}$
    \n#### b) Limit as x approaches 2 from the left:

• Function: f(x) = \begin{cases} 3x - 4 & x \leq 2 \
  3x + 4 & x > 2 \end{cases}

• Evaluation:

  • Find the limit from the left side of 2: $\lim_{x \to 2^-} (3x - 4) = 3(2) - 4 = 6 - 4 = 2$

• Therefore: $\lim_{x \to 2^-} f(x) = 2$
  \n#### c) Limit as x approaches 3:

• Expression: $\lim_{x \to 3} (\sqrt{x} - 3)$

• Justification: Direct substitution gives: $\sqrt{3} - 3$

• Therefore: $\sqrt{3} - 3$
  \n#### d) Limit as x approaches 0 from the right:

• Expression: $\lim_{x \to 0^+} \frac{2x^2 - 3}{x^2}$

• Justification: Direct substitution gives: $\frac{2(0)^2 - 3}{(0)^2} = \frac{-3}{0}$

• Conclusion: The limit approaches negative infinity: $-\infty$
  \n## Derivative Calculation

Problem 2: Find the derivative using the definition

• Function: $f(x) = \frac{x}{x + 1}$

• Find $f'(1)$ using the derivative definition:

  • $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

• Calculate:

  • $f(1) = \frac{1}{1 + 1} = \frac{1}{2}$

  • $f(1 + h) = \frac{1 + h}{2 + h}$

• Therefore:

  • $f'(1) = \lim_{h \to 0} \frac{\frac{1 + h}{2 + h} - \frac{1}{2}}{h}$
    \n## Tangent Line Calculation

Problem 3: Find the equation of the tangent line

• Function: $f(x) = x^3 - 3x^2 + x + 4$

• Point of tangency: (3,7)

• Find the derivative:

  • $f'(x) = 3x^2 - 6x + 1$

• Evaluate at x = 3:

  • $f'(3) = 3(3^2) - 6(3) + 1 = 27 - 18 + 1 = 10$

• Point-slope form:

  • $y - f(3) = f'(3)(x - 3)$

• Substitute:

  • $y - 7 = 10(x - 3)$

• Simplified equation:

  • $y = 10x - 30 + 7$

  • $y = 10x - 23$
    \n## Derivative of Functions

Problem 4: Find the derivatives (no simplification required)

a) Function: $f(x) = x^4 - 2x \sqrt{x}$

• Derivative: $f'(x) = 4x^3 - 2(\sqrt{x} + \frac{2x}{2\sqrt{x}}) = 4x^3 - 2\sqrt{x}(1 + \frac{2}{\sqrt{x}})$
  \n#### b) Function: $g(t) = 5\cdot 2t^3$

• Derivative: $g'(t) = 5 \cdot 6t^2 = 30t^2$
  \n#### c) Function: $y = x\sqrt{6x^2 + 10}$

• Derivative:

  • Use product rule and chain rule:

  • $\frac{dy}{dx} = \sqrt{6x^2 + 10} + x \cdot \frac{1}{2\sqrt{6x^2 + 10}}(12x)$
    \n#### d) Implicit differentiation for: $2xy = y^2 - 3x + 1$

• Differentiate both sides:

  • $2y + 2x\frac{dy}{dx} = 2y\frac{dy}{dx} - 3$

• Solve for $\frac{dy}{dx}$ at the point (2,5):

  • Rearranging gives:

  • $(2x - 2y)\frac{dy}{dx} = -2y + 3$

  • Hence at (2,5):

  • $\frac{dy}{dx} = \frac{1}{2}$

  • Resulting in the final evaluation of $\frac{dy}{dx}$ evaluated at the point (2,5):

  • $\frac{dy}{dx} = -- \frac{2y - 3}{2x - 2y}$