Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

1. Limits and Continuity

Finding Limits Algebraically

Calculus

1. Limits and Continuity

Finding Limits Algebraically

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Multiple Choice

How many tangent lines to the curve $y = \frac{x}{x + 1}$ pass through the point $(1, \frac{1}{2})$ ?

1

0

2

3

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Verified step by step guidance

Step 1: Begin by finding the derivative of the given curve y = x/(x + 1) to determine the slope of the tangent line at any point on the curve. Use the quotient rule: $ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} $, where $ u = x $ and $ v = x + 1 $.

Step 2: Apply the quotient rule to compute $ \frac{dy}{dx} $. Substitute $ u = x $, $ u' = 1 $, $ v = x + 1 $, and $ v' = 1 $ into the formula: $ \frac{dy}{dx} = \frac{(1)(x + 1) - (x)(1)}{(x + 1)^2} $. Simplify the expression to find the slope of the tangent line at any point $ x $.

Step 3: Write the equation of the tangent line in point-slope form: $ y - y_1 = m(x - x_1) $, where $ m $ is the slope $ \frac{dy}{dx} $, and $ (x_1, y_1) $ is a point on the curve. Substitute $ y_1 = \frac{x_1}{x_1 + 1} $ (from the curve equation) and $ m = \frac{dy}{dx} $.

Step 4: Determine the condition for the tangent line to pass through the external point (1, 1/2). Substitute $ x = 1 $ and $ y = 1/2 $ into the tangent line equation. Solve for $ x_1 $, which represents the x-coordinate of the point of tangency on the curve.

Step 5: Solve the resulting equation for $ x_1 $. This will yield the x-coordinates of the points on the curve where the tangent lines pass through (1, 1/2). Verify that there are two distinct solutions for $ x_1 $, confirming that there are two tangent lines.