Originally Posted by
Chris L T521 Sure.
$\displaystyle f(x)=\left\{\begin{array}{rl}x^2&x<0\\2&x=0\\2x+1& x>0\end{array}\right.$
I'll evaluate $\displaystyle f(-2)$
Since $\displaystyle -2<0$, you want to focus on the function that is defined for $\displaystyle x<0$. Looking at the piecewise function, we see something:
$\displaystyle f(x)=\left\{\begin{array}{rl}{\color{red}x^2}&{\co lor{red}x<0}\\2&x=0\\2x+1&x>0\end{array}\right.$
We see that the function defined for $\displaystyle x<0$ is $\displaystyle x^2$. So we can now say that $\displaystyle f(-2)=(-2)^2=\color{red}\boxed{4}$
Does this clarify things? Can you try the others on your own?
--Chris