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Chapter 6: Problem 42

Let \(f(x)=\left\\{\begin{array}{c}x^{2}\left|\cos \frac{\pi}{x}\right|, \quad x \neq 0 \\ 0, \quad x=0\end{array}, x \in \mathbb{R}\right.\), then \(f\) is (A) differentiable both at \(x=0\) and at \(x=2\) (B) differentiable at \(x=0\) but not differentiable at \(x=2\) (C) not differentiable at \(x=0\) but differentiable at \(x=2\) (D) differentiable neither at \(x=0\) nor at \(x=2\)

Short Answer

Expert verified
The function f is differentiable both at x=0 and at x=2.

Step by step solution

01

Examine the function at x=0

Check if function f is differentiable at x = 0 by testing if the limit as x approaches 0 of (f(x) - f(0)) / (x - 0) exists. We need to compute the following limit: \(\frac{x^2 |\cos(\pi/x)| - 0}{x - 0} = x |\cos(\pi/x)|\). Since \(|\cos(\pi/x)|\) is bounded between 0 and 1, and as x approaches 0, x approaches 0 as well, the product approaches 0. Hence, the function is differentiable at x = 0 as the limit exists and is equal to 0.
02

Find the derivative for non-zero x

Differentiate the function f with respect to x for x not equal to 0. Using the product rule, and the chain rule, the derivative will be: \(f'(x) = 2x |\cos(\pi/x)| + x^2 \frac{d}{dx}(|\cos(\pi/x)|)\). We do not have to compute the exact form of the second part because we are only interested in examining the differentiability at x = 2.
03

Examine the function at x=2

Substitute x = 2 into the derivative to see if it's defined, which would imply differentiability at that point. When x = 2, \(f'(2) = 4 |\cos(\frac{\pi}{2})| + 4 \frac{d}{dx}(|\cos(\frac{\pi}{2})|)\). Since \(\cos(\frac{\pi}{2})\) is 0, the modulus is also 0, and thus its derivative will be 0. Therefore, there is no issue with differentiability at x = 2.

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Key Concepts

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

Limit of a Function
Understanding the limit of a function is crucial for studying calculus. A limit is a value that a function approaches as the input approaches some value. Limits are essential in defining both continuity and differentiability.

For instance, when we say the limit of a function like \(f(x)\) as \(x\) approaches zero is \(L\), written as \(\lim_{x \to 0} f(x) = L\), it means that we can make \(f(x)\) as close as we want to \(L\), by making \(x\) sufficiently close to zero. This concept is vital when determining differentiability at a point, as the derivative itself is defined via limits.
Product Rule for Differentiation
The product rule for differentiation is a method used when you need to find the derivative of a product of two functions. If you have two functions \(u(x)\) and \(v(x)\), the derivative of their product \(u(x)v(x)\) is \(u'(x)v(x) + u(x)v'(x)\).

This rule is often employed in calculus, as products of functions appear frequently. It simplifies calculating the derivative of a product without having to expand the product into a sum of terms first. In our exercise, this rule was applied to differentiate \(x^2 \times |\cos(\pi/x)|\).
Chain Rule for Differentiation
The chain rule is another fundamental tool in calculus. It is used for differentiating composite functions, that is, functions made up of two or more functions. The rule can be summarized as follows: if you have a function \(h(x) = f(g(x))\), then the derivative of \(h\) with respect to \(x\) is \(h'(x) = f'(g(x)) \cdot g'(x)\).

This allows us to 'chain' together the derivatives of the outer function \(f\) and the inner function \(g\). When computing the derivative of \(|\cos(\frac{\pi}{x})|\), for example, one would use the chain rule to differentiate the outer modulus function and the inner cosine function.
Continuity and Differentiability
Continuity and differentiability are closely related concepts in calculus. A function is continuous at a point if it’s defined at that point, and the limit of the function as it approaches that point is the same as the function's value. Differentiability, on the other hand, means that a function has a defined derivative at that point.

To be differentiable at a point, a function must be continuous there. However, continuity does not guarantee differentiability. As observed in the solution, to examine differentiability at \(x = 0\), we checked if the derivative limit existed. Since it did, the function was differentiable at that point.
Properties of Cosine Function
The cosine function is a trigonometric function that has several important properties. It is periodic, with a period of \(2\pi\), meaning that \(\cos(x + 2\pi) = \cos(x)\) for all values of \(x\). It is also an even function, which means that \(\cos(-x) = \cos(x)\).

In the context of the exercise, the absolute value of the cosine function, \(|\cos(x)|\), was used. The absolute value affects the analysis for differentiability because it can introduce points where the function is not differentiable. However, at \(x = 2\), this was not problematic because \(\cos(\pi/2) = 0\), and therefore the absolute value did not introduce any sharp kinks or corners.

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Most popular questions from this chapter

A proton is fired from very far away towards a nucleus with charge \(Q=120 e\), where \(e\) is the electronic charge. It makes a closest approach of \(10 \mathrm{fm}\) to the nucleus. The de Broglie wavelength (in units of \(\mathrm{fm}\) ) of the proton at its start is: (take the proton mass, \(m_{p}=(5 / 3) \times 10^{-27} \mathrm{~kg} ;\) \(\left.\mathrm{h} / e=4.2 \times 10^{-15} \mathrm{~J} . \mathrm{s} / \mathrm{C} ; \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~m} / \mathrm{F} ; 1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)\)

A mixture of 2 moles of helium gas (atomic mass \(=4 \mathrm{amu}\) ) and 1 mole of argon gas (atomic mass \(=40 \mathrm{amu}\) ) is kept at \(300 \mathrm{~K}\) in a container. The ratio of the \(\mathrm{rms}\) speeds \(\left(\frac{v_{n n s}(\text { helium })}{v_{r m s}(\text { argon })}\right)\) is (A) \(0.32\) (B) \(0.45\) (C) \(2.24\) (D) \(3.16\)

Identify the binary mixture(s) that can be separated into individual compounds, by differential extraction, as shown in the given scheme. (A) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (B) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\) (C) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) (D) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\)

If \(y(x)\) satisfies the differential equation \(y^{\prime}-y \tan x=2 x \sec x\) and \(y(0)=0\), then (A) \(y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}\) (B) \(y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{18}\) (C) \(y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9}\) (D) \(y^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}\)

Let \(P=\left[a_{i j}\right]\) be a \(3 \times 3\) matrix and let \(Q=\left[b_{i j}\right]\), where \(b_{i j}=2^{i+j} a_{i j}\) for \(1 \leq i, j \leq 3\). If the determinant of \(P\) is 2 , then the determinant of the matrix \(Q\) is (A) \(2^{10}\) (B) \(2^{11}\) (C) \(2^{12}\) (D) \(2^{13}\)
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