Graph each function. Give the domain and range. ƒ(x) = (1/3)

Question

Graph each function. Give the domain and range. ƒ(x) = (1/3)^-x+1

Accepted Answer

Hello everybody. Uh We're doing all right. Today, today we're going to be looking at this question that states provide the domain and range of the function by graphing it where our function is A X is equal to open parentheses, one divided by 11 closed parentheses to the exponent negative X plus three. Now when it comes to a function like this, the first thing I want to try to do is see if there's any simple, simple, more simple form of this equation. So what I mean by that is to see if there's a parent function, right? So I know I see that my F of X has different qualities that I want to get rid of that I can achieve the most basic version of that function. And in this case, I'm gonna label my parent function as GEO X. And I'll say that G of X is equal to open parentheses, one divided by 11, close parentheses to the exponent X. So that would be the most basic way of writing our function. We haven't added any additional parts to it yet. Now this function, so G of X has the form of open parentheses one divided by a, where A is a constant could be any or, or it could be any number to the exponent X. So open parentheses, one divided by a close parentheses to the exponent X. And that graph, we usually, I'll draw a very basic form of that graph. So I draw a Y axis and an X axis that graph will have a Y intercept at A Y and it will have a horizontal ascent tilt at Y equals zero. So the X axis will function as a horizontal ascent tode so past the Y intercept will be approaching the Y axis or the X axis but never touching it. And then extending infinitely to the right and to the left of the Y axis will be going infinitely upwards. So it's kind of in the shape of an L or uh you can think of it as a stretched L L as an and an L as a curve, a curved L. So that's a general form of a graph of one divided by eight to the exponent X. Now we have to see how we jump from our parent function G of X to our current function E and to do that, we'll go little by little, basically just adding different parts of our function until we arrive to the Effex. So first, I'll say that we have another function H of X. And I'll first add one of the qualities of our, of of our F of X function. So I'll say that F of X is equal to open parentheses, one divided by 11 closed parentheses. And first, I'll add the negative in front of the X. So I'll say to the exponent negative XX. And this means that G of X or the graph of geo reflex across the Y axis. So that is my first change to my parent function and to the graph of my parent function. Now we'll jump to another change and I'll label this change, this will be our F of X. So this is how we achieve our F of X. So our F of X is equal to open parentheses, one divided by 11 closed parentheses to the exponent negative X. And now finally, we add the plus three to the exponent. And that means that our H H of X shifts or the graph of H O X shifts three units, right? So now I see how I can go from my parent function. I obtain my parent function and a general graph of what my parent function would look like. And I've obtained what transformations I do to that paragraph to achieve my F of X function. So now I'm going to try different points in my F of X. So let's say F of three. And that means I'm plugging in three for my X. So I'll have F of three as equal to open parentheses. One divided by 11 closed parentheses to the exponent negative three plus three, which means we're going to have an exponent of zero and anything to the exponent zero is one. So we'll have that F of three is equal to one or the 0.3 comma one. So that's our first point here and then we can try a different X. So I'll say fa four equals open parentheses, one divided by 11, close parentheses to exponent negative four plus three. And now we'll have uh a negative exponent. So ne on the exponent, we have negative four plus three, which is negative one. And then we have negative one as the exponent specifically, especially with a fraction. We're using the reciprocal of that fraction. So we'll have that F of four is equal to 11 or the 0.4 comma 11. So now I have two different points and I see how I move from my parent graph to my current function. So now we will graph F O X. So to do that once again, we'll draw a Y axis and an X axis and we'll label our points. So we had a point at an X of three, another point at an ex of four. And as for our Y values, we said that we had a, a point at A Y of one and another point at A Y of 11 and our point at a Y of one, so that was three comma one. That point makes sense because if you think about it, we had a Y intercept first at one comma zero. And now if we shift that three units to the right, as we mentioned, then that Y intercept will move and it's no longer a Y intercept. Now it'll be at a point of three comma one. So the other point that we had was four comma 11. So we'll put a point there as well. We'll connect those two points as a curve and we'll extend past it in infinitely upwards and to the right. And before the three comma one, our horizontal Asymptote continues to be the X axis. All we did was reflect across the y axis. So the horizontal Asymptote does not change in this case. So we'll extend to the left approaching the X axis but never touching it and will send infinitely left. So now our graph should look like an extended L. So a curved L but a reflected curved L since as we mentioned, we reflect our graph across the Y axis. So now we can detect the domain and the range. And remember domain has to do with the possible X values in our graph. So now we have to look at the shape of our graph. The shape of a graph extends to the left. It extends infinitely extends infinitely to the left and to the right. Although it extends slower at a slower rate to the right, it could still extends infinitely to the right. So our domain will be from negative infinity to positive infinity. Since we can, since we can have any possible value of X left or right now, our range has to do with possible Y values of our graph. So we have to see how it changes up and down the Y axis. Now, we know that we have a horizontal ascent at A Y of zero. So that means that we don't go below the X axis and we also never touch the X axis. So our range will be an open parenthesis and it's an open parenthesis because we never touch the Y axis or the X axis. If we did, it would be a bracket. So we put up open parentheses from zero, comma zero is our starting point here. And then we see how the graphic stands upwards and we see that ex extends infinitely upwards. So it'll be from zero to positive infinity and close parentheses. And once again, because we never touch positive infinity, we'll have a parentheses, closing out our range. If we were to touch the end point of our range, it would be a bracket. So just like that, we have identified the graph of our function F of X as well as identifying the domain arrange to go along with it. So I hope that was helpful. And until next time

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