1. ## Piecewise-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

2. Originally Posted by magentarita
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
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

3. ## ok...

Originally Posted by Chris 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
It does make slight sense but can you do at least one of them for me?

4. Originally Posted by magentarita
It does make slight sense but can you do at least one of them for me?
Sure.

$f(x)=\left\{\begin{array}{rl}x^2&x<0\\2&x=0\\2x+1& x>0\end{array}\right.$

I'll evaluate $f(-2)$

Since $-2<0$, you want to focus on the function that is defined for $x<0$. Looking at the piecewise function, we see something:

$f(x)=\left\{\begin{array}{rl}{\color{red}x^2}&{\co lor{red}x<0}\\2&x=0\\2x+1&x>0\end{array}\right.$

We see that the function defined for $x<0$ is $x^2$. So we can now say that $f(-2)=(-2)^2=\color{red}\boxed{4}$

Does this clarify things? Can you try the others on your own?

--Chris

5. ## ok.....

Originally Posted by Chris L T521
Sure.

$f(x)=\left\{\begin{array}{rl}x^2&x<0\\2&x=0\\2x+1& x>0\end{array}\right.$

I'll evaluate $f(-2)$

Since $-2<0$, you want to focus on the function that is defined for $x<0$. Looking at the piecewise function, we see something:

$f(x)=\left\{\begin{array}{rl}{\color{red}x^2}&{\co lor{red}x<0}\\2&x=0\\2x+1&x>0\end{array}\right.$

We see that the function defined for $x<0$ is $x^2$. So we can now say that $f(-2)=(-2)^2=\color{red}\boxed{4}$

Does this clarify things? Can you try the others on your own?

--Chris
Thank you so much.