Question 61, Find the range of $\mathrm{f}(\mathrm{x})=\mathrm{x}+\sqrt{5-\mathrm{x}^{4}}$.

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Question 61, Solution Method One: Clearly, $x \in[-\sqrt[4]{5}, \sqrt[4]{5}]$.
When $x \in[-\sqrt[4]{5}, 0]$, $f(x)$ is monotonically increasing, so $-\sqrt[4]{5} \leq f(x) \leq \sqrt{5}$.
When $x \in[0, \sqrt[4]{5}]$, for any real number $a$, according to the Cauchy-Schwarz inequality, we have:
$$
\begin{array}{l}
\left(x+\sqrt{5-x^{4}}\right)^{2}=\left(\frac{1}{a} \cdot a x+1 \cdot \sqrt{5-x^{4}}\right)^{2} \leq\left(\frac{1}{a^{2}}+1\right)\left(a^{2} x^{2}+5-x^{4}\right) \\
=\left(\frac{1}{a^{2}}+1\right)\left[-\left(x^{2}-\frac{a^{2}}{2}\right)^{2}+\frac{a^{4}}{4}+5\right] \leq\left(\frac{1}{a^{2}}+1\right)\left(\frac{a^{4}}{4}+5\right)
\end{array}
$$

The above inequality holds if and only if: $a^{4} x^{2}=5-x^{4}$, and $x^{2}=\frac{a^{2}}{2}$. Solving these, we get $a=\sqrt{2}, x=1$. Hence, $x+\sqrt{5-x^{4}} \leq \sqrt{\left(\frac{1}{2}+1\right)(1+5)}=3$.
In summary, the range of $f(x)$ is $[-\sqrt[4]{5}, 3]$.
Solution Method Two: Clearly, $x \in[-\sqrt[4]{5}, \sqrt[4]{5}]$.
Differentiating, we get $f(x)=1-\frac{2 x^{3}}{\sqrt{5-x^{4}}}$. Clearly, when $x \in(-\sqrt[4]{5}, 1)$, $f(x)>0$, i.e., $f(x)$ is monotonically increasing; and when $x \in(1, \sqrt[4]{5})$, $f'(x)<0$, i.e., $f(x)$ is monotonically decreasing. Also, $f(-\sqrt[4]{5})=-\sqrt[4]{5}$, $f(1)=3$, and $f(\sqrt[4]{5})=\sqrt[4]{5}$. Therefore, the range of $f(x)$ is $[-\sqrt[4]{5}, 3]$.
Solution Method Three: Clearly, $x \in[-\sqrt[4]{5}, \sqrt[4]{5}]$.
When $x \in[-\sqrt[4]{5}, 0]$, $f(x)$ is monotonically increasing, so $-\sqrt[4]{5} \leq f(x) \leq \sqrt{5}$.
When $x \in[0, \sqrt[4]{5}]$, let $y=\sqrt{5-x^{4}}$, then $y^{2}+x^{4}=5$. We need to find the maximum value of $x+y$. Let $F(x, y, \lambda)=x+y-\lambda\left(x^{4}+y^{2}-5\right)$. According to the method of Lagrange multipliers, the critical points satisfy:

Thus, $f_{\text{max}}(x)=f(1)=3$.
In summary, the range of $f(x)$ is $[-\sqrt[4]{5}, 3]$.