Question:
The interval in which the function \( f(x) = 4x^2 + x \) is decreasing is:
The interval in which the function \( f(x) = 4x^2 + x \) is decreasing is:
Show Hint
To determine where a function is increasing or decreasing:
- Compute the first derivative \( f'(x) \).
- Solve \( f'(x)>0 \) for increasing intervals.
- Solve \( f'(x)<0 \) for decreasing intervals.
- Ensure to check critical points and denominators for undefined regions.
Updated On: Apr 14, 2026
- \( \left( -\frac{1}{2}, \frac{1}{2} \right) \)
- \( \left[ -\frac{1}{2}, \frac{1}{2} \right] \)
- \( (-1, 1) \)
- \( [-1, 1] \)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Compute the first derivative \( f'(x) \).
Given:
\[
f(x) = \frac{4x^2 + 1}{x}
\]
Using the quotient rule:
\[
\left( \frac{g(x)}{h(x)} \right)' = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}
\]
where \( g(x) = 4x^2 + 1 \) and \( h(x) = x \), we compute their derivatives:
\[
g'(x) = 8x, \quad h'(x) = 1
\]
Applying the quotient rule:
\[
f'(x) = \frac{(8x \cdot x) - (4x^2 + 1) \cdot 1}{x^2}
\]
\[
f'(x) = \frac{8x^2 - 4x^2 - 1}{x^2}
\]
\[
f'(x) = \frac{4x^2 - 1}{x^2}
\]
Step 2: Find where \( f'(x) \) is negative.
\[
\frac{4x^2 - 1}{x^2}<0
\]
Since \( x^2 \) in the denominator is always positive, the inequality simplifies to:
\[
4x^2 - 1<0
\]
\[
4x^2<1
\]
\[
x^2<\frac{1}{4}
\]
\[
-\frac{1}{2}<x<\frac{1}{2}
\]
Thus, \( f(x) \) is decreasing in the interval:
\[
\left( -\frac{1}{2}, \frac{1}{2} \right)
\]
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