For the piecewise-defined below, find:

f(-2), f(0) and f(2)

.........{x^2..........if x < 0

f(x) = {2..............if x = 0

.........{2x + 1.......if x > 0

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- Oct 29th 2008, 07:48 AMmagentaritaPiecewise-defined Function
For the piecewise-defined below, find:

f(-2), f(0) and f(2)

.........{x^2..........if x < 0

f(x) = {2..............if x = 0

.........{2x + 1.......if x > 0 - Oct 29th 2008, 07:51 AMChris L T521
Which of these piecewise functions is defined for negative values of x (i.e. x<0)? Substitute -2 into that function to get the value of f(-2).

There is only one function that is defined at x=0. So what do you think f(0) would be?

Which of these piecewise functions is defined for positive values of x (i.e. x>0)? Substitute 2 into that function to get the value of f(2).

Does this make sense?

--Chris - Oct 29th 2008, 07:58 AMmagentaritaok...
- Oct 29th 2008, 08:05 AMChris 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 - Oct 29th 2008, 06:11 PMmagentaritaok.....