If \(\mathrm{f}(x)=\left(\sin ^4 x+\cos ^4 x\right), 0 \lt x \lt \frac{\pi}{2}\), then the function has minimum value _______ at \(x=\) ________.

\(\begin{aligned}
f(x) & =\sin ^4 x+\cos ^4 x \\
& =\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x \\
& =1-\frac{1}{2}(\sin 2 x)^2
\end{aligned}\)
Since \(0 \leq \sin ^2 2 x \leq 1\)
\(\therefore \quad 0 \geq-\frac{1}{2} \sin ^2 2 x \geq-\frac{1}{2}\)
\(\begin{aligned} & \Rightarrow 1+0 \geq 1-\frac{1}{2} \sin ^2 2 x \geq 1-\frac{1}{2} \\ & \Rightarrow 1 \geq \sin ^4 x+\cos ^4 x \geq \frac{1}{2}\end{aligned}\)
\(\begin{aligned} & \Rightarrow 1-\frac{1}{2}(\sin 2 x)^2=\frac{1}{2} \\ & \Rightarrow(\sin 2 x)^2=1 \\ & \Rightarrow(\sin 2 x)^2=\left(\sin \frac{\pi}{2}\right)^2 \\ & \Rightarrow 2 x=\frac{\pi}{2} \\ & \Rightarrow x=\frac{\pi}{4}\end{aligned}\)