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Magazine Chapter 23: Problem 31
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 \( f'(x) = \frac{-2}{x^2} \), and the function is differentiable for \( x \neq 0 \).
Step by step solution
01
Understand the Function
The given function is \( y = 1 + \frac{2}{x} \). This is composed of a constant, \(1\), and a term that involves division by \(x\).
02
Definition of Derivative
The derivative of a function \( f(x) \) at a point \(x\) is given by the limit definition: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]. Here, we have to apply this to \( f(x) = 1 + \frac{2}{x} \).
03
Find \( f(x+h) \)
Compute \( f(x+h) \). Substitute \(x + h\) for \(x\) in \(y = 1 + \frac{2}{x}\):\[ f(x+h) = 1 + \frac{2}{x+h} \].
04
Substitute in Derivative Definition
Using the derivative limit definition, substitute \( f(x) = 1 + \frac{2}{x} \) and \( f(x+h) = 1 + \frac{2}{x+h} \):\[ f'(x) = \lim_{h \to 0} \frac{\left(1 + \frac{2}{x+h}\right) - \left(1 + \frac{2}{x}\right)}{h} \].
05
Simplify the Expression
Cancel the \(1\)'s and simplify the fraction:\[ f'(x) = \lim_{h \to 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \]. Combine the fractions in the numerator:\[ f'(x) = \lim_{h \to 0} \frac{2(x) - 2(x+h)}{h(x)(x+h)} \]. This simplifies to:\[ f'(x) = \lim_{h \to 0} \frac{-2h}{h(x^2 + xh)} \].
06
Cancel \( h \) and Evaluate the Limit
Cancel out \(h\) from the numerator and denominator:\[ f'(x) = \lim_{h \to 0} \frac{-2}{x^2 + xh} \]. As \(h\) approaches 0, the expression becomes:\[ f'(x) = \frac{-2}{x^2} \].
07
Determine Differentiability
The function \( y = 1 + \frac{2}{x} \) has a derivative wherever the denominator in the derivative is not zero. Since the denominator is \(x^2\), the derivative is undefined only at \(x = 0\). Thus, the function is differentiable for \( x eq 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
The derivative is a concept from calculus that represents the rate at which a function is changing at any given point. It is essentially the slope of the tangent line to the function at a particular point. To find the derivative of a function, we frequently use the limit definition, which is key in calculus.
- The limit definition of the derivative is: \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).
- This formula computes how much the function \( f(x) \) changes as we make a tiny step \( h \) away from \( x \).
- In our example, the function is \( y = 1 + \frac{2}{x} \), so to find \( f'(x) \), we applied this definition using our specific function.
Differentiability
Differentiability is a property of a function that tells us whether it has a derivative at each point in its domain. For a function to be differentiable at a point, its derivative must exist at that point. This usually means the function must be smooth and continuous there. However, differentiability is more strict than continuity; a function must fulfill the conditions of differentiability to have its derivative.
- In the example \( y = 1 + \frac{2}{x} \), differentiability was analyzed through the derivative we found: \( f'(x) = \frac{-2}{x^2} \).
- The function is not differentiable at points where the denominator of the derivative is zero since division by zero is undefined.
- For this function, \( x^2 \) in the denominator means the function is not differentiable at \( x = 0 \).
Limits
Limits are a foundational concept of calculus, allowing us to define derivatives, integrals, and continuity. A limit describes the value that a function approaches as the input approaches some value. In the context of derivatives, limits help us find the exact rate of change of the function.
- The formula \( \lim_{h \to 0} \) in the derivative definition shows how we calculate the slope of the function as \( h \) gets infinitesimally small.
- Limits process small intervals and help understand behavior near specific points which might not be exactly computable otherwise.
- They simplify expressions causing potential complex change patterns, as seen when solving \( \frac{-2h}{h(x^2 + xh)} \) into \( \frac{-2}{x^2} \).
Rational Functions
Rational functions are quotients of two polynomials, offering interesting properties and challenges, especially when it comes to calculus. The function in our exercise \( y = 1 + \frac{2}{x} \) can be seen as a rational function since it incorporates division by a variable.
- Rational functions present unique behaviors like asymptotes, which can occur if the polynomial in the denominator equals zero, causing discontinuity or points of non-differentiability.
- In our example, \( \frac{2}{x} \) means the function approaches infinity or undefined when \( x = 0 \), making the function non-differentiable at that point.
- Understanding when a rational function is differentiable or continuous requires careful analysis of these asymptotic behaviors, often using derivatives to illustrate their impact.
