Question

The maximum value of the function \[ y = \frac{x^2}{x^4 + 4}, x \in \mathbb{R}, \] is ............

Hint:

For rational functions, critical points often occur where numerator or denominator derivatives balance — check both for extremum.

Answer 1

Correct Answer: 0.25

Step 1: Differentiate $y$.
\[ y' = \frac{(2x)(x^4 + 4) - x^2(4x^3)}{(x^4 + 4)^2} = \frac{2x(x^4 + 4 - 2x^4)}{(x^4 + 4)^2} = \frac{2x(4 - x^4)}{(x^4 + 4)^2}. \]

Step 2: Set derivative to zero.
\[ 2x(4 - x^4) = 0 \implies x = 0 \text{ or } x^4 = 4 \implies x = \pm \sqrt{2}. \]

Step 3: Evaluate $y$ at critical points.
At $x = 0$: $y = 0$. At $x = \sqrt{2}$: $y = \dfrac{2}{4 + 4} = \dfrac{1}{4}$.

Step 4: Conclusion.
\[ \boxed{\text{Maximum value is } \frac{1}{4}.} \]