Piecewise-defined Function

**magentarita** · October 29th 2008, 07:48 AM

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

**Chris L T521** · October 29th 2008, 07:51 AM

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

**magentarita** · October 29th 2008, 07:58 AM

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?

**Chris L T521** · October 29th 2008, 08:05 AM

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

**magentarita** · October 29th 2008, 06:11 PM

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.

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