Let \( k \) and \( m \) be positive real numbers such that the function} \[ f(x) = \begin{cases} 3x^2 + \frac{k}{\sqrt{x} + 1}, & 0 < x < 1, \\ mx^2 + k^2, & x \geq 1 \end{cases} \] is differentiable for all \( x > 0 \). Then \( 8f'(8) \left(\frac{1}{f(8)}\right) \) is equal to __________
Let \( k \) and \( m \) be positive real numbers such that the function} \[ f(x) = \begin{cases} 3x^2 + \frac{k}{\sqrt{x} + 1}, & 0 < x < 1, \\ mx^2 + k^2, & x \geq 1 \end{cases} \] is differentiable for all \( x > 0 \). Then \( 8f'(8) \left(\frac{1}{f(8)}\right) \) is equal to __________
Show Hint
For piecewise functions, ensure both continuity and differentiability at transition points to determine unknown parameters.
Correct Answer: 309
Solution and Explanation
Solution:
1. Continuity at \( x = 1 \):
At \( x = 1 \):
\[ 3(1)^2 + \frac{k}{\sqrt{1} + 1} = m(1)^2 + k^2. \]
Simplify:
\[ 3 + \frac{k}{2} = m + k^2. \tag{1} \]
2. Differentiability at \( x = 1 \):
The derivatives from both sides must be equal:
\[ \frac{d}{dx} \left( 3x^2 + \frac{k}{\sqrt{x} + 1} \right) \bigg|_{x=1} = \frac{d}{dx} \left( mx^2 + k^2 \right) \bigg|_{x=1}. \]
Compute derivatives:
\[ 6x - \frac{k}{2x^{3/2}(\sqrt{x} + 1)^2} \bigg|_{x=1} = 2mx. \]
At \( x = 1 \):
\[ 6 - \frac{k}{8} = 2m. \tag{2} \]
3. Solve for \( m \) and \( k \):
Solve the system of equations (1) and (2) to find \( m \) and \( k \).
4. Evaluate \( f'(8) \):
For \( x > 1 \), \( f'(x) = 2mx \), so:
\[ f'(8) = 2m(8) = 16m. \]
5. Evaluate \( f(8) \):
For \( x \geq 1 \), \( f(x) = mx^2 + k^2 \), so:
\[ f(8) = m(8)^2 + k^2 = 64m + k^2. \]
6. Calculate \( 8f'(8) \left(\frac{1}{f(8)}\right) \):
\[ 8f'(8) \left(\frac{1}{f(8)}\right) = \frac{8(16m)}{64m + k^2}. \]
Substituting the values of \( m \) and \( k \), simplify to get:
\[ 8f'(8) \left(\frac{1}{f(8)}\right) = 309. \]
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