Problem 29 Find the derivative of each func... [FREE SOLUTION]

Chapter 23: Problem 29

Find the derivative of each function by using the definition. Then determine the values for which the function is differentiable. $$y=1+\frac{2}{x}$$

Short Answer

Expert verified

The derivative is $\frac{-2}{x^2}$, and the function is differentiable for all $x\neq 0$.

Step by step solution

Understand the Definition of a Derivative

The derivative of a function at a point $x$ is defined as $ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} $. For the function $ y = 1 + \frac{2}{x} $, we need to apply this definition to find the derivative.

Set Up the Limit Expression

The function is $ y = f(x) = 1 + \frac{2}{x} $. We first determine $ f(x+h) = 1 + \frac{2}{x+h} $. Substitute $ f(x) = 1 + \frac{2}{x} $ and $ f(x+h) $ into the definition of a derivative: \[ f'(x) = \lim_{{h \to 0}} \frac{(1 + \frac{2}{x+h}) - (1 + \frac{2}{x})}{h} \]}},{

Simplify the Expression

After canceling the constants, we have: \[ f'(x) = \lim_{{h \to 0}} \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \] Find a common denominator for the fractions inside the limit: \[ f'(x) = \lim_{{h \to 0}} \frac{\frac{2x - 2(x + h)}{x(x+h)}}{h} \] Simplify the numerator: \[ f'(x) = \lim_{{h \to 0}} \frac{\frac{-2h}{x(x+h)}}{h} \] This simplifies to: \[ f'(x) = \lim_{{h \to 0}} \frac{-2}{x(x+h)} \] Finally, take the limit as $ h \to 0 $ to get the derivative: \[ f'(x) = \frac{-2}{x^2} \]

Determine Where the Function is Differentiable

A function is differentiable wherever its derivative exists. Notably, a term with $ x $ in the denominator requires $ x eq 0 $, as division by zero is undefined. Therefore, the function $ y = 1 + \frac{2}{x} $ is differentiable for all $ x $ except $ x = 0 $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Definition of Derivative

The limit definition of the derivative is a cornerstone in calculus. It describes how we determine the slope of a function at a specific point. Consider the formula:

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

This represents the function's instantaneous rate of change at $ x $. It gives us the steepness or slope of the tangent.

For $ y = 1 + \frac{2}{x} $, the method involves evaluating limits:

First, find an expression for $ f(x + h) $.
Substitute into the limit definition.
Simplify the expression.
Finally, evaluate the limit as $ h \to 0 $.

Through these steps, you'll find the derivative as $ f'(x) = \frac{-2}{x^2} $. This expression tells us about the function's behavior and slope at different points.

Differentiability

Differentiability is a property that tells us where a function has a derivative. If a derivative exists at a point, the function is said to be differentiable at that point.

If the derivative does not exist, the function is not differentiable there.
Differentiability implies a smooth, non-vertical tangent at that point on the curve.
If a graph has breaks, sharp turns, or vertical tangents, it fails to be differentiable at those points.

By applying these concepts to our function $ y = 1 + \frac{2}{x} $, notice: at $ x = 0 $, the function approaches infinity. This results in a non-existent derivative, rendering the function non-differentiable at that point. For all other values of $ x $, $ f'(x) = \frac{-2}{x^2} $ exists, thus the function is differentiable on its domain except where division by zero occurs.

Rational Functions

Rational functions are expressions that involve a ratio of polynomials. They're known for producing curves characterized by asymptotic behavior and complex domain boundaries. When dealing with these functions, it's helpful to consider:

The expressions often contain fractions, which dictates strict attention to the function's domain.
They can exhibit vertical asymptotes where the function value heads toward infinity; typically at solutions where the denominator is zero.
Rational functions can be differentiable excluding points where discontinuous.

In $ y = 1 + \frac{2}{x} $, note how the term $ \frac{2}{x} $ affects its behavior. Understanding its domain, $ x eq 0 $, is essential for determining where differentiation is possible. The derivative $ f'(x) = \frac{-2}{x^2} $ highlights how vertical asymptotes influence the graph and its derivative's existence.

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91影视

Find the derivative of each function by using the definition. Then determine the values for which the function is differentiable. $$y=1+\frac{2}{x}$$

Short Answer

Step by step solution

Understand the Definition of a Derivative

Set Up the Limit Expression

Simplify the Expression

Determine Where the Function is Differentiable

Key Concepts

Limit Definition of Derivative

Differentiability

Rational Functions

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