How would you graph this piccewise function??? { -x+1 if x0 } f(x)={ -3x if xor equal to 3 }

Question

How would you graph this piccewise function???
        {  -x+1 if x>0 }
f(x)={  -3x    if x<3 }
        {   5       if x>or equal to 3 }

ranga · Answer

Are you sure the domain is typed correctly?  The first one has x > 0 and the second one says x < 3.   If i am interested in x =2 do I use -x + 1  or -3x?

anonymous · Answer

Medal to who ever answers this.

ranga · Answer

My question has not been answered yet. Can you verify it says -x + 1 if x > 0 and -3x if x < 3 When x = 2, x > 0 and I should use the first equation. But when x = 2, x < 3 and I should I use the second equation. Which is it? That is why I am asking are you sure there are no typos?

anonymous · Answer

NOPE.

anonymous · Answer

where did u get the x=2 anyways?

ranga · Answer

Just a point I chose between 0 and 3 because it falls under both domains: It is greater than 0 and so it is the first domain and it is less than 3 and therefore falls in the second domain.

ranga · Answer

If the first domain is x < 0 (that is, "less than" instead of "greater than" ) then it would make sense.

anonymous · Answer

I'm supposed to graph the points so wouldn't the first one be (0,1) with a slope of -1/1

ranga · Answer

Then it is not a function because the same x value will give two different results.  It will fail the vertical line test.

But this can still be plotted, except it won't be a function.

anonymous · Answer

That's what it says on my worksheet so I guess nothing's wrong.

ranga · Answer

f(x) = -x + 1  for  x > 0

This will be a straight line, downward sloping, inclined at -45 degrees, cutting the x-axis at x = 1.  The line should not touch or cross the y-axis because this is defined only for x > 0.

ranga · Answer

The above line should have an open circle at x = 0 f(x) = -3x for x < 3 This is also a downward sloping straight line, that passes through the origin. The line should not go past x = 3 because its domain is x < 3. The line should have an open circle at x = 3 f(x) = 5 for x >= 3 This is a horizontal line, at height (or y-value 3). This line should not go less than 3 because its domain is x >=3. It should have a closed circle at x = 3.

anonymous · Answer

Ok thanks I got it now! (: