34 The maximum value of the function $y=x\left(1+\sqrt{1-x^{2}}\right)$ is $\qquad$ , and the minimum value is $\qquad$ .
$34 y_{\text {max }}=\frac{3 \sqrt{3}}{4}, y_{\text {min }}=-\frac{3 \sqrt{3}}{4}$. Because
$$
\begin{aligned}
y^{2} & =x^{2}\left(1+\sqrt{1-x^{2}}\right)^{2}=\left[1-\left(1-x^{2}\right)\right]\left(1+\sqrt{1-x^{2}}\right)^{2} \\
& =\left(1-\sqrt{1-x^{2}}\right)\left(1+\sqrt{1-x^{2}}\right)^{3} \\
& \leqslant \frac{1}{3} \cdot\left[\frac{3\left(1-\sqrt{1-x^{2}}\right)+3\left(1+\sqrt{1-x^{2}}\right)}{4}\right]^{4} \\
& =\frac{27}{16},
\end{aligned}
$$

Therefore, $|y| \leqslant \frac{3 \sqrt{3}}{4}$. When $x=\frac{\sqrt{3}}{2}$, $y$ reaches its maximum value, and when $x=-\frac{\sqrt{3}}{2}$, $y$ reaches its minimum value.