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Chapter 10: Problem 28

Suppose that \(f(x)=x g(x)\) for some function \(g\) which is continuous at \(0 .\) Prove that \(f\) is differentiable at \(0,\) and find \(f^{\prime}(0)\) in terms of \(g .\)

Short Answer

Expert verified
Since \( g \) is continuous at \( 0 \), \( f'(0) = g(0) \).

Step by step solution

01

- Understanding the Problem

Given the function \( f(x) = x g(x) \), where \( g \) is continuous at \( x = 0 \). We need to prove that \( f \) is differentiable at \( x = 0 \) and find \( f'(0) \).
02

- Definition of Differentiability

A function \( f(x) \) is differentiable at \( x = 0 \) if the following limit exists: \[ f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} \].
03

- Evaluate \( f(0) \)

Substitute \( x = 0 \) into the function \( f(x) \): \[ f(0) = 0 \cdot g(0) = 0 \].
04

- Evaluate \( f(0 + h) \)

Substitute \( x = h \) into the function \( f(x) \): \[ f(h) = h g(h) \].
05

- Apply Definition of Differentiability

Use the definition from Step 2 and substitute the expressions from Steps 3 and 4: \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h g(h) - 0}{h} = \lim_{h \to 0} g(h) \].
06

- Use Continuity of \( g \)

Since \( g \) is continuous at \( x = 0 \), \( \lim_{h \to 0} g(h) = g(0) \). Thus, \[ f'(0) = g(0) \].

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Key Concepts

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

Limits
In calculus, the concept of a limit is essential. It describes the behavior of a function as its input approaches a certain value. For example, the limit of a function as the input approaches zero is a way to understand how the function behaves near that specific point. Mathematically, it's expressed as \( \lim_{h \to 0} f(h) \) for some function \(f\). In the exercise, we use the limit to determine if a function is differentiable.
Continuity
Continuity of a function means that there are no interruptions in its graph. Specifically, a function \[ g(x) \] is continuous at \( x = 0 \) if \( \lim_{x \to 0} g(x) = g(0) \). This property is pivotal for proving differentiability. In the given exercise, the continuity of \( g(x) \) at \( x = 0 \) is exploited to show that \( \lim_{h \to 0} g(h) \) is equal to \( g(0) \), reinforcing the behavior of the function near that point.
Derivative
The derivative of a function at a point provides the rate at which the function is changing at that point. It's the foundation for many concepts in calculus. For a function \( f(x) \), the derivative at \( x = 0 \) is defined by \[ f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} \]. Through the steps given, we evaluate this limit for the function \( f(x) = x g(x) \). By substituting and simplifying, we find that \( f'(0) = g(0) \), demonstrating the importance of the limit in identifying the derivative.
Calculus
Calculus is the branch of mathematics that studies continuous change. It comprises two main parts: differential calculus and integral calculus. Differential calculus deals with the concept of a derivative, allowing us to find how a function changes at any given point. The exercise provided uses differential calculus to determine \( f'(0) \) for the function \( f(x) = x g(x) \), where the core idea is identifying the rate of change of \(f \) using limits and the properties of continuity of \(g \). By linking all these concepts, we see the power of calculus in solving complex problems efficiently.

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Most popular questions from this chapter

Give an example of functions \(f\) and \(g\) such that \(g\) takes on all values, and \(f \circ g\) and \(g\) are differentiable, but \(f\) isn't differentiable. (The problem becomes trivial if we don't require that \(g\) takes on all values; \(g\) could just be a constant function, or a function that only takes on values in some interval \((a, b),\) in which case the behavior of \(f\) outside of \((a, b)\) would be irrelevant.)

(a) If \(f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0},\) find a function \(g\) such that \(g^{\prime}=f\) Find another. (b) If $$f(x)=\frac{b_{2}}{x^{2}}+\frac{b_{3}}{x^{3}}+\dots+\frac{b_{m}}{x^{m}}$$ find a function \(g\) with \(g^{\prime}=f\) (c) Is there a function$$f(x)=a_{n} x^{n}+\cdots+a_{0}+\frac{b_{1}}{x}+\cdots+\frac{b_{m}}{x^{m}}$$such that \(f^{\prime}(x)=1 / x ?\)

(a) If \(\left.g=f^{2} \text { find a formula for } g^{\prime} \text { (involving } f^{\prime}\right)\) (b) If \(\left.g=\left(f^{\prime}\right)^{2}, \text { find a formula for } g^{\prime} \text { (involving } f^{\prime \prime}\right)\) (c) Suppose that the function \(f>0\) has the property that $$\left(f^{\prime}\right)^{2}=f+\frac{1}{f^{3}}$$ Find a formula for \(f^{\prime \prime}\) in terms of \(f .\) (In addition to simple calculations, a bit of care is needed at one point.)

(a) Prove that if \(f\) is differentiable at \(a,\) then \(|f|\) is also differentiable at a provided that \(f(a) \neq 0\) (b) Give a counterexample if \(f(a)=0\) (c) Prove that if \(f\) and \(g\) are differentiable at \(a\), then the functions \(\max (f, g)\) and \(\min (f, g)\) are differentiable at \(a,\) provided that \(f(a) \neq\) \(g(a)\) (d) Give a counterexample if \(f(a)=g(a)\)

In Leibnizian notation the Chain Rule ought to read: $$\frac{d f(g(x))}{d x}=\left.\frac{d f(y)}{d y}\right|_{y=g(x)} \cdot \frac{d g(x)}{d x}$$ Instead, one usually finds the following statement: "Let \(y=g(x)\) and \(z=f(y) .\) Then $$\frac{d z}{d x}=\frac{d z}{d y} \cdot \frac{d y}{d x}$$ Notice that the \(z\) in \(d z / d x\) denotes the composite function \(f \circ g,\) while the \(z\) in \(d z / d y\) denotes the function \(f ;\) it is also understood that \(d z / d y\) will be "an expression involving \(y, "\) and that in the final answer \(g(x)\) must be substituted for \(y .\) In each of the following cases, find \(d z / d x\) by using this formula; then compare with Problem 1. (i) \(\quad z=\sin y, \quad y=x+x^{2}\) (ii) \(\quad z=\sin y, \quad y=\cos x\) (iii) \(\quad z=\sin u, \quad u=\sin x\) (iv) \(\quad z=\sin v, \quad v=\cos u, \quad u=\sin x\)
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