China Mathematical Competition

Let $a^x=y$. The original function is then changed to $$g(y)=y^2+3y-2,$$ which is increasing over $\left(-\frac{3}{2},+\infty \right)$.

When $0<a<1$, we have $y\in [a,a^{-1}]$ and $$g(y{)}_{\max }=a^{-2}+3a^{-1}-2=8\Rightarrow a^{-1}=2\Rightarrow a=\frac{1}{2}.$$ Then $$g(y{)}_{\min }={\left(\frac{1}{2}\right)}^2+3\times \frac{1}{2}-2=-\frac{1}{4}.$$ When $a>1$, we have $y\in [a^{-1},a]$ and $$g(y{)}_{\max }=a^2+3a-2=8\Rightarrow a=2.$$ Then $$g(y{)}_{\min }=2^{-2}+3\times 2^{-1}-2=-\frac{1}{4}.$$ In summary, the minimum value of $f(x)$ on $x\in [-1,1]$ is $-\frac{1}{4}$.