 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Domain Restrictions and Functions Defined Piecewise - Problem 3

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Let's graph another piecewise defined function. This time h(x) equals 2 to the x minus 3 for x less than or equal to 2. Root x minus 2 plus 4 for x greater than 2. Let's get to know the function by computing some values first.

H(0), 0 falls into this category. It's less than or equal to 2. So h(0) is going to be 2 to the 0 minus 3, which is 1 minus 3, or -2. H(2); well, 2 is still in this interval. So I'll get 2 to the 2 minus 3, that's 4 minus 3 or 1. Now 2.01 falls into this interval. It's greater than 2. So I have the square root of 2.01 minus 2. The square root of .01 plus 4. That's not as bad as it looks. .01 is .1², so the square root is .1. So that's 4.1. Then h(6). 6 is also greater than 2, so I have root 6 minus 2, root 4 which is 2 plus 4, 6.

Let's graph this. First I need a set of axis. Let's think about the two shapes that we're going to be graphing. You've got 2 to the x minus 3, that's a transformation of y equals 2 to the x which looks like this. 2 to the x minus 3 is a downward shift of this graph. Remember that this graph has a horizontal asymptote at y equals 0. So if you shift it down three units, the asymptote will also shift down 1, 2, 3 down to here. So the whole graph shifts down.

So the first thing I want to note is that asymptote. That's going to be for this piece of the graph which goes up to x equals 2. So let me draw the asymptote at y equals -3. Then I'm going to have that go to x equals 2. So let's call this 2. The asymptote doesn't have to stop. Remember it's just a guideline for your graph.

Now the 2 to the x minus 3 part. Remember that 2 to the 0 is 1, so 1 minus 3 is -2. So we'll have a point here. 2 to the 1 is 2 minus 3 is -1. We do go up to x equals 2. 2 to the 2 we already calculated is 4 minus 3 is 1, so we have a negative point here at y equals 1. So our left piece is going to look something like this.

What about our right piece? Our right piece is a transformed square root graph. That's going to look something like this. Let's put it over here. The square root graph starts here. It's little endpoint is at 0,0 to begin this. But after we shift it to the right 2 units, and up 4 units. It goes over here 2 units to the right 1, 2, 3, 4 units up. We're going to have this square root graph up here. So I need to find the point 2 units to the right, and 4 units up that's where my square root graph is going to begin.

So that's two units to the right 1, 2, 3, 4 units up. That's where the square root graph begins. I don't really want a closed circle here, because keep in mind this part of the graph is only defined for x greater than 2. So I'll have an open circle here instead. The shape's going to be the same. We've only just translated the square root graph. So I can draw a square root graph which looks something like this. That's our final graph.

Now let's answer the question; what is the domain, and what is the range. The domain you can read right off of the definition of the function. It's defined for x less than or equal to 2, and for x greater than 2. That's basically all the numbers. All numbers are either greater than 2, or less than or equal to 2. So the domain is all real numbers.

What about the range? This is interesting, because if you look at the graph here, the set of y values that we can get from our graph, it goes up to 1, and down 2, but not including -3. That's what this is. So that's going to give us the interval from -3 to 1, not including -3. -3 to 1. We also have this piece here which starts at 4 a, d just goes up. It does not include 4. So from 4 onto infinity. That's our domain, and range and this is our graph of y equals h of x.