Graph each function. Give the domain and range. ƒ(x) = log￬1/2 (x...

Question

Graph each function. Give the domain and range. ƒ(x) = log￬1/2 (x-2)

Accepted Answer

Hello everybody. I am doing all right today that they are going to be looking at this question that states consider the logarithmic function F of X is equal to the log of base 1/8 of open parentheses, X minus three, closed parentheses and graphic and identify its domain and range. Now for a question like this, where we have our function, I want to identify if there's another parent function that this is based off of. So in this case, my parent function, I'll write it as G of X is equal to and my parent function in this case is just log of 1/8 of X or a log of base 1/8 of X. So I'll have that G of X is equal to log of base 1/8 of X. And the function F of X is just a horizontal translation of this, of G of X three units to the right. So that's where the X minus three comes in. So the minus three means that we're translating the graph of G of X three units to the right. So now that I have that I can try to plug some points in So I'm going to try to first graph the G of X my parent function. And then once I have that graph, I can just translate that graph three units to the right. And that will be the graph of our F of X function. So let's do A T chart, let's make A T chart with two columns, X values on the left and Y values on the right where we can try different XS as 0.5 12, 34 and five. And all we do is I'll make an example here. So if we have G of one half or G of 0.5 that would be equal to log base 1/8 of one half and that is equal to one third. So to make those calculations, you can, you can just use your calculator where you plug in log of base 1/8 of one half. So we have our first value. So if our X is 0.5 our Y will be one third and you can just continue to do that for the other Xes. So if we have X of one, we have Y equals zero because the log of any base of one is zero. Now, if we plug in too far, X, we'll have a Y of negative one third. If we plug in three, we'll have a Y of negative 0.53. If we plug in four for the X, we'll have a Y of negative two thirds. And if we plug in five, we'll have a Y of negative 0.77 and we can turn this into points. So our first point will be 0.5 comma one third. Our second point will be one comma zero. So that will be our, what? Our X intercept, our third point will be two comma negative one third. Our fourth point will be three comma negative 0.53. Our fifth point will be four comma negative two thirds. And our final point will be five comma negative 0.77. So another note or something else that we should take into account is that the graph of G of X, let's say any function equals log of base A of X has a vertical, ask him to at the Y axis. So for any graph that has the form log base A of X, there will be an Asymptote at the Y axis. So first we'll do a graph of G of X, our G of X function, as we said, we'll have an Asy tode at the Y axis. So we'll always be very close to it but never touching. It will have an X intercept at an X of one and at an X of 0.5 will be at one third. So it will be below one. So as we come down from the approaching the Y axis, we come down and once we pass one, we start making a turn since we know at zero and an X of 0.5 will be at a Y of one third and then at an X of one will be at a Y of zero. We know we're turning downwards and then that graph will continue to head downwards and expand that way. Now, if we turn that graph or translate it three units to the right to have our F of X graph, which is the graph that we want for this question. Once again, we'll draw a Y axis and then X axis and we can do kind of the same thing what we did with our G of X where we just plug in certain points that we know we have. So this will be the same graph graph as G of X but three units to the right. So instead of having a point at one comma zero, we'll have a point at four comma zero. So X intercept will be at four instead of one. And because we had a vertical Asymptote at the Y axis, if we move that three units to the right, that ascent to three units to the right, our Asymptote will now be at an X of three. So we'll have an Aceto, a vertical aceto A X equals three. So the graph once again, as we come down at X equal three, we'll be approaching Equus three but never touching it. Once we pla pass A Y of one, we know we're gonna have to start turning downwards to intersect the Y, the X axis at an X of four. And then that will continue to be pass to pass the X axis and continue to go downwards turning as well. And that will extend always upward and always downward. So that will be our graph for our function F of X. So now we can determine the domain and the range. So the domain has to do with possible X values of our graph. In this case, we'll have an open parenthesis. And then let's look at the possible X value. So our graph, our F of X graph extends has a vertical Asymptote at X equals three. So we don't go anywhere left of X equals three because that is a vertical asthma, we never pass it. So our domain will start at three and will extend while it got our graph, our graph extends infinitely to the right, which is positive X. So go from three comma two positive infinity and we close parentheses and that is our domain. So now we look at the range, the range on the other hand has to do with possible Y values. So we see that we have a vertical Asymptote, but we don't have any horizontal asy tode. And we see that our graph extends infinitely upwards and infinitely downwards. Therefore, our range will be an open parentheses from negative infinity to positive infinity since any Y values will be possible in our graph. And just like that, we have obtained a graph of our function F of X as well as its domain and its range. So I hope that was helpful and I'll see you next time.

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Graph each function. Give the domain and range. ƒ(x) = log￬1/2 (x-2)

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