Question

If \( f(x) = \begin{cases} x^3 \sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases} \), then:

• A. \( f''(0) = 1 \)
• B. \( f''\left(\frac{2}{\pi}\right) = \frac{24 - \pi^2}{2\pi} \)
• C. \( f''\left(\frac{2}{\pi}\right) = \frac{12 - \pi^2}{2\pi} \)
• D. \( f''(0) = 0 \)

Answer 1

The Correct Option is B

The given function is:

\( f(x) = x^3 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right). \)

The second derivative of \( f(x) \) is computed as:

\( f''(x) = 6x \sin\left(\frac{1}{x}\right) - 3 \cos\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) + \sin\left(\frac{1}{x}\right) \left(-\cos\left(\frac{1}{x}\right)\right). \)

Substitute \( x = \frac{2}{\pi} \):

\( f''\left(\frac{2}{\pi}\right) = 6\left(\frac{2}{\pi}\right) \sin\left(\frac{\pi}{2}\right) - 3 \cos\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{2}\right). \)

Simplify:

\( f''\left(\frac{2}{\pi}\right) = \frac{12}{\pi} - \frac{\pi^2}{2\pi} = \frac{24 - \pi^2}{2\pi}. \)

Thus, the final value is:

\( f''\left(\frac{2}{\pi}\right) = \frac{24 - \pi^2}{2\pi}. \)

Answer 2

To solve this problem, we need to determine the values of the second derivative \( f''(x) \) for the given piecewise function \( f(x) \).

The given function is:

\(f(x) = \begin{cases} x^3 \sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}\)

To proceed, we need to compute the derivatives of \( f(x) \). Let's start with the first derivative \( f'(x) \) for \( x \neq 0 \):

\(f'(x) = \frac{d}{dx} \left(x^3 \sin\left(\frac{1}{x}\right)\right)\)

Using the product rule, \(\frac{d}{dx}(uv) = u'v + uv'\), where \( u = x^3 \) and \( v = \sin\left(\frac{1}{x}\right) \):

1. Compute \(u' = \frac{d}{dx}(x^3) = 3x^2\).
2. Compute \(v' = \frac{d}{dx}(\sin\left(\frac{1}{x}\right)) = -\cos\left(\frac{1}{x}\right)\cdot\frac{d}{dx}\left(\frac{1}{x}\right) = \frac{\cos\left(\frac{1}{x}\right)}{x^2}\).

Using the product rule,

\(f'(x) = 3x^2 \sin\left(\frac{1}{x}\right) + x^3 \cdot \left(-\frac{\cos\left(\frac{1}{x}\right)}{x^2}\right)\)

This simplifies to:

\(f'(x) = 3x^2 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right)\)

Now we find the second derivative \( f''(x) \) for \( x \neq 0 \):

\(f''(x) = \frac{d}{dx}\left(3x^2 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right)\right)\)

1. Derivative of \(3x^2 \sin\left(\frac{1}{x}\right)\) is:
   • \(\frac{d}{dx}(3x^2) = 6x\)
   • So, \(u'v = 6x \sin\left(\frac{1}{x}\right)\)
   • And, \(uv' = 3x^2 \cdot \left(-\frac{\cos\left(\frac{1}{x}\right)}{x^2}\right) = - 3 \cos\left(\frac{1}{x}\right)\)
2. Derivative of \(-x \cos\left(\frac{1}{x}\right)\) is:
   • \(-\cos\left(\frac{1}{x}\right)\)
   • And \(-x \cdot \left(-\frac{\sin\left(\frac{1}{x}\right)}{x^2}\right) = \frac{\sin\left(\frac{1}{x}\right)}{x}\)

Therefore,

\(f''(x) = 6x \sin\left(\frac{1}{x}\right) - 3 \cos\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) + \frac{\sin\left(\frac{1}{x}\right)}{x}\)

Simplifying further gives:

\(f''(x) = 6x \sin\left(\frac{1}{x}\right) + \frac{\sin\left(\frac{1}{x}\right)}{x} - 4 \cos\left(\frac{1}{x}\right)\)

Now, evaluate at \( x = \frac{2}{\pi} \):

\(f''\left(\frac{2}{\pi}\right) = 6 \cdot \frac{2}{\pi} \sin\left(\frac{\pi}{2}\right) + \frac{1}{\frac{2}{\pi}} \sin\left(\frac{\pi}{2}\right) - 4\cos\left(\frac{\pi}{2}\right)\)

Since \(\sin\left(\frac{\pi}{2}\right) = 1\) and \(\cos\left(\frac{\pi}{2}\right) = 0\), we have:

\(f''\left(\frac{2}{\pi}\right) = \frac{12}{\pi} + \frac{\pi}{2} - 0\)

Thus, simplifying yields:

\(f''\left(\frac{2}{\pi}\right) = \frac{24 - \pi^2}{2\pi}\)

The correct answer is \(f''\left(\frac{2}{\pi}\right) = \frac{24 - \pi^2}{2\pi}\).