Chapter 10: Problem 20
Let \(f: \mathrm{R} \rightarrow \mathrm{R}\) be defined as \(f(x)=\left\\{\begin{array}{cc}x^{5} \sin \left(\frac{1}{x}\right)+5 x^{2}, & x<0 \\ 0, & x=0 \\ x^{5} \cos \left(\frac{1}{x}\right)+\lambda x^{2}, & x>0\end{array}\right.\) The value of \(\lambda\) for which \(f^{\prime \prime}(0)\) exists, is
Short Answer
Step by step solution
Recognize the piecewise function
Calculate the first derivative of f(x)
Evaluate derivative continuity at x=0
Calculate the second derivative of f(x)
Evaluate continuity of second derivative at x=0
Conclusion: Verify calculation consistency
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piecewise function
In the given exercise, the function \(f(x)\) is defined piecewise like this:
- For \(x < 0\): \(f(x) = x^5 \sin\left(\frac{1}{x}\right) + 5x^2\). This part involves a mix of polynomial and trigonometric behavior.
- For \(x = 0\): \(f(x) = 0\). This is a constant value for the point at \(x = 0\).
- For \(x > 0\): \(f(x) = x^5 \cos\left(\frac{1}{x}\right) + \lambda x^2\). Here, \(\lambda\) is an unknown parameter we need to determine.
Continuity
For a piecewise function to be fully continuous:
- The function value must be consistent across all pieces at transition points (like at \(x=0\)).
- The derivatives (first and second) should also transition smoothly between different pieces.
When \(f'(x)\) is examined, it was shown that limits from both sides as \(x\) approaches 0 (\(x \to 0^-\) and \(x \to 0^+\)) were equal to zero. To ensure this, \(2\lambda x\) and \(10x\) needed to match when approaching zero, leading to setting \(\lambda = 5\).
The same value of \(\lambda\) ensures the second derivative \(f''(x)\) is continuous, with both sides of the limit meeting at the value 10 when \(x \to 0\). Every step is essential to guarantee continuity at \(x=0\).
Derivative
- First Derivative \(f'(x)\): This involves using the product and chain rules to differentiate each segment of \(f(x)\). For \(x < 0\), the derivative was found to include terms like \(5x^4 \sin\left(\frac{1}{x}\right)\), and for \(x > 0\), included \(2\lambda x\).
- Second Derivative \(f''(x)\): We differentiated \(f'(x)\), resulting in terms like \(20x^3 \sin\left(\frac{1}{x}\right)\). Important at \(x=0\) was that the result needed to match for limits from both sides.