7. Let $f(x)=\sin ^{4} x-\sin x \cos x+\cos ^{4} x$, then the range of $f(x)$ is $\qquad$
$$
\begin{array}{l}
f(x)=\sin ^{4} x-\sin x \cos x+\cos ^{4} x \\
=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x-\sin x \cos x=-2 \sin ^{2} x \cos ^{2} x-\sin x \cos x+1 \\
\xlongequal{t=\sin x \cos x \in\left[-\frac{1}{2}, \frac{1}{2}\right]}-2 t^{2}-t+1=-2\left(t+\frac{1}{4}\right)^{2}+\frac{9}{8} \in\left[0, \frac{9}{8}\right]
\end{array}
$$

The translation is as follows:

$$
\begin{array}{l}
f(x)=\sin ^{4} x-\sin x \cos x+\cos ^{4} x \\
=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x-\sin x \cos x=-2 \sin ^{2} x \cos ^{2} x-\sin x \cos x+1 \\
\xlongequal{t=\sin x \cos x \in\left[-\frac{1}{2}, \frac{1}{2}\right]}-2 t^{2}-t+1=-2\left(t+\frac{1}{4}\right)^{2}+\frac{9}{8} \in\left[0, \frac{9}{8}\right]
\end{array}
$$