Characteristics and Domain and Range of Exponential Functions

Why study the exponential function?

The exponential function is used in real life to model certain types of growth and decay. Let's think about a  simple example to explore the idea.

Consider a house, valued at $\$50000$$50000 that is set to increase in value by $10$10% each year. In the first year, the house's value will become $\$550000$$550000 (determined by the calculation $500000+10%\left(500000\right)=550000$500000+10%(500000)=550000). In the second year, the house's value becomes $\$605000$$605000, which is an increase of $\$55000$$55000 over the previous year.

So even though the house increased by the same $10$10% each year, the actual amount of increase was more in the second year than that of the first year. This is a key idea. The rate of growth in dollars in the house price is proportional to the house price itself. Less expensive houses will grow in value much slower than expensive houses.  

Alternatively If an item depreciates (loses value) by $10$10% each year, then the largest drop in value will occur in the first year. As the years go on, the amount of fall in dollar value will accordingly reduce. It will never reduce to zero because we are always calculating $10$10% of something.

The basic mathematical model for this type of process has the independent variable in the exponent. Typically, the model has the general form $y=Ab^{mx+c}+k$y=Abmx+c+k, The number $b$b is called the base of the function and the other constants ($A$A, $m$m $c$c, and $k$k) are there to allow the model to match the particular real life growth or decay rate under consideration.  

For example, without giving the details of the calculation, the house price example can be modelled by the function $y=500000\times10^{0.041393x}$y=500000×100.041393x where $x$x is the number of years of growth (note that $A=500000$A=500000, the base $b=10$b=10 and the growth rate $m$m is the fraction $0.41393$0.41393). The point being that we can use the function to determine future values of the house. This is the power of models.

Domain and Range

Like any function, we are interested in knowing the domain and range of the exponential function.

Remember that the domain is the set of values that the independent variable (usually $x$x) can take, and the range is the set of values that the dependent variable (usually $y$y) can take. 

Unless there are restrictions imposed on the function, the natural domain of the general exponential function $y=Ab^{mx+c}+k$y=Abmx+c+k is the entire set of real numbers. So for functions like $y=2^x$y=2x, $y=3^{-x}$y=3−x, $y=\left(0.5\right)^x+3$y=(0.5)x+3, $y=5-10^x$y=5−10x etc all have domains given by $x\in\Re$x∈ℜ. 

This means that any real value of $x$x can be substituted into any exponential function of this form to calculate  an associated $y$y value.  For example, at $x=0$x=0, the function $y=2^x$y=2x becomes $2^0=1$20=1, and so the y-intercept of the function $y=2^x$y=2x is $1$1.

The $y$y-intercept of the general function $y=Ab^{mx+c}+k$y=Abmx+c+k is determined by setting $x=0$x=0. So for $y=\left(0.5\right)^x+3$y=(0.5)x+3, the $y$y intercept is $y=\left(0.5\right)^0+3=1+3=4$y=(0.5)0+3=1+3=4. 

The range of the function is always restricted to one of two forms. For the real number $L$L, the range of an exponential function is given by either $y>L$y>L  or $yy<L. We can think of $L$L as an upper or lower limit for the function, so that the line $y=L$y=L will always be an asymptote.

For example, the range of $y=2^x$y=2x and $y=2^{-x}$y=2−x are both given by $y>0$y>0 and the range of $y=8-2^{-x}$y=8−2−x is given by $y<8$y<8. 

 

Examining two graphs  

We will examine two simple graphs to illustrate some features of the exponential function. 

Graph 1

The first is a graph of $y=2^{-x}$y=2−x shown here.

Starting from the left, we see the graph falling quickly. For example, although some not shown, the $y$y values of $x=-3$x=−3, $x=-2$x=−2 and $x=-1$x=−1 are $2^{-\left(-3\right)}=2^3=8$2−(−3)=23=8, $2^{-\left(-2\right)}=2^2=4$2−(−2)=22=4 and $2^{-\left(-1\right)}=2^1=2$2−(−1)=21=2.  We might say that the rate of change of $y$y is severe, dropping from $8$8 to $4$4 to $2$2 over an $x$x interval of $3$3 units.

The $y$y-intercept is $1$1, and this would be true if the graph were $y=3^{-x}$y=3−x, or $y=4^{-x}$y=4−x or in fact more generally $y=b^{-x}$y=b−x.   

As we proceed further, the curve gradually slows its rate of descent, but nevertheless always falling toward the x axis. It continues to become closer and closer to it, but never reaches it.

We call the line to which a curve is approaching but never reaches an asymptote. This means that the $x$x-axis, which has the equation $y=0$y=0 is an asymptote for the function. Mathematically, we say that as $x$x approaches infinity, the function approaches zero. In mathematical terms, we write as $x\rightarrow\infty,y\rightarrow0$x→∞,y→0. 

It is clear from the graph that the range of the function clearly includes all positive reals. This means we can find the value of x that produces any particular value of y. For example, for a value like $y=\frac{1}{4}$y=14​, we simply form the equation $2^{-x}=\frac{1}{4}$2−x=14​, we rewrite as $2^{-x}=2^{-2}$2−x=2−2, and thus we realise that $x=2$x=2 is mapped to $y=\frac{1}{4}$y=14​.

Graph 2

The second example concerns the exponential  function $y=9-3^x$y=9−3x. Think of this function as the function $y=3^x$y=3x first reflected in the x axis (becoming $y=-3^x$y=−3x and then lifted up by $9$9 units to become $y=9-3^x$y=9−3x. Here is the graph:

The first thing to note is that the asymptote is the line $y=9$y=9 shown on the diagram. Thus as $x\rightarrow-\infty,y\rightarrow9$x→−∞,y→9. This is clear from the function itself, because, as the quantity $3^x$3x is approaching zero, $9-3^x$9−3x must be approaching $9$9. 

The domain of the function is given by $x\in\Re$x∈ℜ and the range is given by $y<9$y<9.

You will also note that three points have been highlighted. The $y$y intercept is found by putting $x=0$x=0 into the function, showing that $9-3^0=8$9−30=8. The $x$x intercept is found by putting $9-3^x=0$9−3x=0, from which we see that $3^x=9$3x=9 and so $x=2$x=2.

Finally, we note that if $x=1$x=1, then $9-3^1=6$9−31=6, and so the point $\left(1,6\right)$(1,6) is on the curve.

Worked Examples

question 1

question 2

question 3