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Chapter 10: Problem 31

Prove that it is impossible to write \(x=f(x) g(x)\) where \(f\) and \(g\) are differentiable and \(f(0)=g(0)=0 .\) Hint: Dillerentiate.

Short Answer

Expert verified
It is impossible to write \( x = f(x) g(x) \) with the given conditions because it leads to a contradiction in the differentiated equation.

Step by step solution

01

Differentiate both sides

Start by differentiating both sides of the equation using the product rule. The original equation is: \[ x = f(x)g(x) \]To differentiate, apply the product rule to the right-hand side: \[ 1 = f'(x) g(x) + f(x) g'(x) \]
02

Substitute given conditions

Next, substitute the given conditions: \( f(0) = 0 \) and \( g(0) = 0 \) into the differentiated equation. At \( x = 0 \), the equation becomes: \[ 1 = f'(0) \times 0 + 0 \times g'(0) \]Simplify this step: \[ 1 = 0 \]
03

Analyze the contradiction

The equation \( 1 = 0 \) obtained in Step 2 is a contradiction. This means the initial assumption that such functions \( f(x) \) and \( g(x) \) exist is incorrect.

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function is changing at any given point. This rate is what we call the derivative. The process of differentiation involves applying specific rules to obtain the derivative of a function.
Product Rule
The product rule is a formula used to find the derivative of a product of two functions. If you have two differentiable functions, say \(f(x)\) and \(g(x)\), the product rule states that the derivative of their product \(f(x)g(x)\) is given by:\[ (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \]This means you take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function.
Contradiction in Proofs
In mathematics, a contradiction occurs when we arrive at an assertion that opposes or contradicts an earlier statement that was assumed or proven to be true. Contradiction is often used in proofs to show that certain assumptions lead to impossible outcomes.

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Most popular questions from this chapter

(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.)

If \(f(x)=x^{-n}\) for \(n\) in \(N,\) prove that $$\begin{aligned}f^{(k)}(x) &=(-1)^{k} \frac{(n+k-1) !}{(n-1) !} x^{-n-k} \\\&=(-1)^{k} k !\left(\begin{array}{c} n+k-1 \\\k\end{array}\right) x^{-n-k}, \quad \text { for } x \neq 0\end{aligned}$$

Particle \(A\) moves along the positive horizontal axis, and particle \(B\) along the graph of \(f(x)=-\sqrt{3} x, x \leq 0 .\) At a certain time, \(A\) is at the point (5,0) and moving with speed 3 units/sec; and \(B\) is at a distance of 3 units from the origin and moving with speed 4 units/sec. At what rate is the distance between \(A\) and \(B\) changing?

Find \(f^{\prime}(0)\) if $$f(x)=\left\\{\begin{array}{ll}g(x) \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.$$ and $$g(0)=g^{\prime}(0)=0$$

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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