Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 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
- 10. Physics Applications of Integrals 2h 22m
2. Intro to Derivatives
Tangent Lines and Derivatives
Problem 3.72
Textbook Question
Are there any points on the curve y = x - 1/(2x) where the slope is 2? If so, find them.

1
To find the points on the curve where the slope is 2, we need to first determine the derivative of the function y = x - \(\frac{1}{2x}\). The derivative, y', represents the slope of the tangent line at any point on the curve.
Differentiate the function y = x - \(\frac{1}{2x}\) with respect to x. The derivative of x is 1, and the derivative of \(-\frac{1}{2x}\) can be found using the power rule and chain rule. Rewrite \(-\frac{1}{2x}\) as \(-\frac{1}{2}x^{-1}\) and differentiate.
The derivative of \(-\frac{1}{2}x^{-1}\) is \(\frac{1}{2}x^{-2}\) or \(\frac{1}{2x^2}\). Therefore, the derivative of the function y = x - \(\frac{1}{2x}\) is y' = 1 + \(\frac{1}{2x^2}\).
Set the derivative equal to 2 to find the x-values where the slope of the tangent is 2: 1 + \(\frac{1}{2x^2}\) = 2.
Solve the equation \(\frac{1}{2x^2}\) = 1 for x. This will give you the x-values where the slope is 2. Substitute these x-values back into the original equation y = x - \(\frac{1}{2x}\) to find the corresponding y-values, thus identifying the points on the curve.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative
The derivative of a function measures the rate at which the function's value changes as its input changes. It is often interpreted as the slope of the tangent line to the curve at a given point. For the curve y = x - 1/(2x), finding the derivative will help determine where the slope equals 2.
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Finding Critical Points
Critical points occur where the derivative of a function is zero or undefined. These points are essential for analyzing the behavior of the function, including identifying local maxima, minima, and points of inflection. In this context, we need to set the derivative equal to 2 to find specific points on the curve.
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Solving Equations
Solving equations involves finding the values of variables that satisfy a given mathematical statement. In this case, after determining the derivative and setting it equal to 2, we will solve for x to find the corresponding y-values on the curve. This process is crucial for identifying the points where the slope of the curve is exactly 2.
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