Subsections

• • Exponential growth
  • Exponential Decay
  
  
• Exercises

Inverse Functions

Consider the functions $f,g$ defined by

$$f(x)=e^x+e^{-x},g(x)=e^x-e^{-x}$$

To be able to get an inverse the function must be one-to-one. You can plot the functions to get a hint as to whether they are invertible or not.

> f:=x->exp(x)+exp(-x);
> plot(f(x),x=-5..5);
> g:=x->exp(x)-exp(-x);
> plot(g(x),x=-5..5);

Both satisfy the vertical-line test but $f(x)$ is not invertible since it does not satisfy the horizontal-line test. Indeed $f$ is not one-to-one, for instance $f(0.5)=f(-0.5)$. From the plot it seems that the function $g$ is one-to-one. In order to determine its inverse we solve for x.

> solve(g(x)=y,x);

We observe that one of the solutions is not defined since the arguement of the logarithm can only be positive. Thus:

$$g^{-1}(y)=ln(\frac{y}{2}+\frac{\sqrt{(y^2+4}}{2})$$

> ginv:=y->ln(y/2+sqrt(y^2+4)/2);

Let's look at the plot along with the line $y=x$ to see if our functions seem to make sense.

> plot({x,g(x),ginv(x)},x=-20..20,y=-20..20,scaling=constrained);

Let's check that we have computed the right inverse. By definition the composotion of the functions should be the line $y=x$ since an inverse is the reflection about this line.

$$(g\circ g^{-1})(y)=y=(g^{-1}\circ g)(x)=x$$

> g(ginv(y));
> simplify(g(ginv(y)));
> ginv(g(x));
> simplify(ginv(g(x)));

We are having difficulty getting $x$ for the last composition. Again, the logarithm can't have a negative arguement and the computer knows this. To find the variable values that will work solve $g(y) > 0$. You may come across this in the exercises.

> solve(g(y)=0,y);

So, you need to let the computer know that the variable will only be $>0$.

Exponential growth

The simple model for growth is exponential growth, where it is assumed that $y'(t)$ is proportional to $y$. That is,

$$\frac{dy}{dt} = ky\quad\hbox{where $k$ is a positive constant.}$$

Separating the variables and integrating (see section 4.4 of the text), we have

$$\int\frac{dy}{y} = \int k dt$$

so that

$$\ln \mid y \mid = kt + C$$

In the case of exponential growth, we can drop the absolute value signs around $y$, because $y$ will always be a positive quantity. Solving for $y$, we obtain

$$\mid y \mid = e^{kt + C}$$

which we may write in the form $y = Ae^{kt}$, where $A$ is an arbitrary positive constant.

Exponential Decay

In a sample of a radioactive material, the rate at which atoms decay is proportional to the amount of material present. That is,

$$\frac{dA}{dt} = -kA$$

where $k>0$ is a constant. This is the same equation as in exponential growth, except that $-k$ replaces $k$. The solution is

$$A(t) = A_0 e^{-kt}$$

where $A_0$ is a positive constant. Physically, $A_0$ is the amount of material present at $t=0$.

Radioactivity is often expressed in terms of an element's half-life. For example, the half-life of carbon-14 is 5730 years. This statement means that for any given sample of ${}^{14}\hbox{C}$, after 5730 years, half of it will have undergone decay. So, if the half-life is of an element Z is $c$ years, it must be that $e^{-kc}=\frac{1}{2}$, so that $kc=\ln 2$ and $k=\frac{\ln 2}{c}$.

Exercises

1. For each function given below

   a

   Plot the function over the interval $0 \leq x \leq 100$. Which functions are not invertible and why?

   b

   Find the inverse of the invertible function(s) and plot the function and its inverse along with the line $y=x$ on the intervals $-3 \leq x \leq 3$.

   c

   Show that you have the correct inverse by using the composite definition.

   
   
   $$f_1(x)=\frac{\ln(x-2)}{x}$$
   
   
   
   
   $$f_2(x)=\frac{\ln(x)+5}{3}$$
   
   
   
   
   $$f_3(x)=\frac{1}{\sqrt{\ln(x)^2+1}}$$
   
   
   
   
   $$f_4(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$
   
   

2. Suppose that the population of a certain bacteria can be modeled by an exponential function. In a particular experiment, the number of bacteria was $10,000$ at $t=0$. Four hours later, the bacteria was $46,000$. Find the value of the growth constant $k$ and use it to predict the number of bacteria that would have been present after $6$ hours.
3. How long would it take for half of 75 grams of a radioactive element to decay if it decays at a yearly rate of 2.2%.