Answer
The IVT guarantees there exists a $c \in(0,1)$ such that $$f(c)=2^{-c^{2}}-c=0$$
Work Step by Step
Let $$f(x)=2^{-x^{2}}-x .$$ Observe that $f$ is continuous on $[0,1]$ with $$f(0)=2^{0}-0=1\gt0$$ and $$f(1)= 2^{-1}-1\lt 0 .$$ Therefore, the IVT guarantees that there exists a $c \in(0,1)$ such that $$f(c)=2^{-c^{2}}-c=0$$
