Question Video: Finding the Domain and Range of Exponential Functions | Nagwa Question Video: Finding the Domain and Range of Exponential Functions | Nagwa

# Question Video: Finding the Domain and Range of Exponential Functions Mathematics • Second Year of Secondary School

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Find the domain and range of the function π(π₯) = 7^(π₯ + 5) + 5 in β.

04:44

### Video Transcript

Find the domain and range of the function π of π₯ equals seven to the power of π₯ plus five plus five in the set of real numbers.

Letβs begin by reminding ourselves what we mean by the domain and range of a function. The domain is essentially the set of possible values that we can input to that function, in other words, the set of π₯-values that we can substitute into the function π of π₯. Then the range is the set of possible outputs from the function after the values from the domain have been substituted in. Now we have in particular an exponential function here. And the most general form of the exponential function is π of π₯ equals π to the power of π₯. π is a real number greater than zero and not equal to one.

So weβre going to begin by identifying what the graph of the function π of π₯ looks like. This will allow us to identify its domain and range. Then we can look at the series of transformations that map π of π₯ onto π of π₯, and that will give us the graph of the function π of π₯. Then we can find its domain and range. In particular, weβre going to begin by sketching the graph of π of π₯ equals seven to the power of π₯.

Since the base seven is a number greater than one, then we know we have exponential growth instead of exponential decay. The graph looks a lot like this. It passes through the π¦-axis at one, and it has a horizontal asymptote given by the line π¦ equals zero, or the π₯-axis. It also passes through the point one, seven. Now we can learn this fact, or we can substitute π₯ equals one into the function π of π₯ and we do indeed get seven.

With that in mind, letβs identify what the domain and range of the function π of π₯ is. The domain is the set of possible values we can input into the function. We can think about this as the spread of values in the π₯-direction. There are no limits here. We can substitute any π₯-value into the function π of π₯ and we will always get a real output. The range is a little bit different though. The range can be thought of as the spread of values in the π¦-direction. And we said that there was a horizontal asymptote given by the line π¦ equals zero. This means the function π of π₯ approaches zero but never quite reaches it. In the other direction though, it heads off towards positive β. And so the range is the open interval from zero to positive β.

With this in mind, how do we map the function π of π₯ equals seven to the power of π₯ onto the function π of π₯? Letβs begin by mapping seven to the power of π₯ onto seven to the power of π₯ plus five. This is a horizontal translation of the function π of π₯ five units to the left or by the vector negative five, zero. Then what happens if we add five to the function seven to the power of π₯ plus five? Well, this is a vertical translation five units up or by the vector zero, five. So to map the function π of π₯ equals seven to the power of π₯ onto the function π of π₯, weβre going to perform two translations, or we can think about it as one big translation. Itβs going to move five units left and then five units up.

This time we know it will have a horizontal asymptote five units above the previous one. Since the previous one was given by the line π¦ equals zero, the new asymptote will have the equation π¦ equals five. And whilst not entirely necessary, we can find the value of the π¦-intercept by substituting π₯ equals zero into the function. When we do, we get seven to the fifth power plus five. And we can also see that this passes through the point one, seven to the power of six plus five. Thatβs found by substituting one into the function.

We now, however, have enough information to determine the domain and range of this new function. Once again, the domain can be thought of as the spread of values in the π₯-direction. This is once again unbounded. So the domain of π of π₯ is the set of real numbers. The range once again can be thought of as the spread of values in the π¦-direction. This time the horizontal asymptote is given by the line π¦ equals five, so π of π₯ approaches five but never quite reaches it. In the other direction, it heads off towards positive β. So the range of the function is the open interval from five to β.

And so by considering graphs and transformations of graphs, weβve determined the domain and range of the function π of π₯ equals seven to the π₯ plus fifth power plus five. The domain is the set of real numbers, and the range is the open interval from five to β.

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