Question:
The maximum value of \( f(x) = \frac{x}{1 + 4x + x^2} \) is
The maximum value of \( f(x) = \frac{x}{1 + 4x + x^2} \) is
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Use derivative tests and critical points to find extrema of rational functions.
Updated On: May 18, 2025
- \( \frac{1}{4} \)
- \( \frac{1}{5} \)
- \( \frac{1}{6} \)
- \( \frac{1}{7} \)
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The Correct Option is C
Solution and Explanation
Let \( f(x) = \frac{x}{x^2 + 4x + 1} \). To find the maximum, take derivative:
\[ f'(x) = \frac{(1)(x^2 + 4x + 1) - x(2x + 4)}{(x^2 + 4x + 1)^2} \] \[ = \frac{x^2 + 4x + 1 - 2x^2 - 4x}{(x^2 + 4x + 1)^2} = \frac{-x^2 + 1}{(x^2 + 4x + 1)^2} \] Setting numerator \( -x^2 + 1 = 0 \Rightarrow x = \pm 1 \).
Substitute \( x = 1 \): \( f(1) = \frac{1}{1 + 4 + 1} = \frac{1}{6} \).
\[ f'(x) = \frac{(1)(x^2 + 4x + 1) - x(2x + 4)}{(x^2 + 4x + 1)^2} \] \[ = \frac{x^2 + 4x + 1 - 2x^2 - 4x}{(x^2 + 4x + 1)^2} = \frac{-x^2 + 1}{(x^2 + 4x + 1)^2} \] Setting numerator \( -x^2 + 1 = 0 \Rightarrow x = \pm 1 \).
Substitute \( x = 1 \): \( f(1) = \frac{1}{1 + 4 + 1} = \frac{1}{6} \).
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