Consider the function $f:[-1,1] \to \mathbb{R}$ defined piecewise by \begin{align*}f(x) = \begin{cases}0: -1 \leq x < 0 \\ g(x): 0\leq x \leq 1 \end{cases}\end{align*} where $g:[0,1] \to \mathbb{R}$ is some function. Suppose $g$ is a non-zero function given by $$g(x) = a_0 + a_1x + a_2x^2 + a_3x^3.$$ Find non-negative real numbers $a_0, a_1, a_2, a_3$ (not all of which are 0) such that $f$ is continuous but not differentiable.
I've been able to find some examples of general continuous, but not differentiable functions, but I have not been able to find one that fits this specific example AND to supply non-negative real numbers that makes it true. My problem is I can't preserve continuity at $f(0)$ but make $f(x)$ non-differentiable given the restriction of the $a_i$'s.
What about $g(x) = x+x^2$? $g(0)=0$ and $g^\prime(0) = 1 \neq 0$ so $f$ is continuous at $0$ but not differentiable at that point.
By the way any $g(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ with $a_0= 0, \ a_1 \neq 0$ and $a_0,a_1,a_2,a_3$ non-negative would work.