Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.72

Chapter 3, Problem 3.72

Are there any points on the curve y = x - 1/(2x) where the slope is 2? If so, find them.

Verified step by step guidance

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.

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.

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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Related Practice

Textbook Question

In Exercises 41–58, find dy/dt.

y = (t tan(t))¹⁰

Textbook Question

Differentiating Implicitly

Use implicit differentiation to find dy/dx in Exercises 1–14.

x⁴ + sin y = x³y²

Textbook Question

For what value or values of the constant m, if any, is

ƒ(x) = { sin 2x, x ≤ 0

{ mx, x > 0

a. continuous at x = 0?

b. differentiable at x = 0?

Give reasons for your answers.

Textbook Question

Derivatives

In Exercises 23–26, find dr/dθ.

r = (1 + sec θ) sin θ

Textbook Question

Assume that functions f and g are differentiable with f(2) = 3, f'(2) = −1, g(2) = −4, and g'(2) = 1. Find an equation of the line perpendicular to the line tangent to the graph of F(x) = (f(x) + 3) / (x − g(x)) at x = 2.

Textbook Question

Suppose that functions ƒ(x) and g(x) and their first derivatives have the following values at x = 0 and x = 1.

x ƒ(x) g(x) ƒ'(x) g'(x)

0 1 1 -3 1/2

1 3 5 1/2 -4

Find the first derivatives of the following combinations at the given value of x.

a. 6ƒ(x) - g(x), x = 1

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