calculus ! please help and if u cannot reject me quick many have

Answer: The behavior of the function can be checked by finding the derivative of it and evaluating it at the two specified points:
[tex]\begin{gathered} f(x)=|x^2-9| \\ \\ \\ \frac{df(x)}{dx}=\frac{|x^2-9|}{x^2-9}(2x) \end{gathered}[/tex]The behaviour at x = -3 and x = 3.
[tex]\begin{gathered} \frac{df(x)}{dx}=\frac{\lvert x^{2}-9\rvert}{x^{2}-9}(2x) \\ \\ \\ x=3\rightarrow\frac{\lvert3^2-9\rvert}{3^2-9}(2\times3)=\frac{0}{0}\times6=0\rightarrow\text{ Not possible} \\ \\ x=-3\rightarrow\frac{\lvert(-3)^2-9\rvert}{(-3)^2-9}(2\times-3)=\frac{0}{0}(-6)\rightarrow\text{ Not possible} \\ \end{gathered}[/tex]The answer therefore is:
[tex]\text{ Option\lparen D\rparen}\rightarrow\text{ Continuous but non-Differentiable }[/tex]Plot for the function:
Not that the f(x) is indeed Differentiable function
, but there can not be any slope at x = -3 and x = 3.