Range of the function

We have: $f(x)=\left\{\begin{array}{l}2(1-x^2)-x-1&,0\le x\le1\\2(x^2-1)-x-1&,1\le x\le\dfrac{3}{2}\end{array}\right.$

Or $\qquad\; f(x)=\left\{\begin{array}{l}-2x^2-x+1&,0\le x<1\\2x^2-x-3&,1\le x\le\dfrac{3}{2}\end{array}\right.$

We have the function $g=-2x^2-x+1$ decreases on $\left[0;1\right)$ and we get $-2\le f(x)\le 1$ with $0\le x<1$ .

Moreover, the function $h=2x^2-x-3$ increases on $\left[1;\dfrac{3}{2}\right]$ , which implies $-2\le f(x)\le 0$ with $1\le x\le\dfrac{3}{2}$ .

So, the range of $f(x)$ on the interval $\left[0;\dfrac{3}{2}\right]$ is $\boxed{-2\le f(x)\le 1}$