Use the point plotting method to sketch the graph of \(f ( x ) = 2^x\) and the graph of \(g ( x ) = \left(\frac{1}{2} \right)^x\) for input values \(x\) between \(- 2\) and \(+ 2\) .

See SolutionThe functions in exercise 3.1.1 are examples of exponential functions.

A basic exponential function is a function of the form \(f ( x ) = b^x\), for \(b\) a positive number different from \(1\).

When \(b > 1\), the function is **increasing** as \(x\) ‘moves’ from left to right, and the function is referred to as an **exponential growth function**.

When \(0 < b < 1\), the function is **decreasing** as \(x\) ‘moves’ from left to right and the function is referred to as an **exponential decay function**.

In either case, the function is a **one-to-one function** so \(b^x=b^z\) if and only if \(x=z\).

The idea of an exponential function is sometimes broadened to include functions of the type
\(f ( x ) = c b^x\) , where \(c\) is a constant different from \(0\). Since \(f ( 0 ) = c\) for such a function, it follows that exponential functions of this type have the form \(f ( x ) = f ( 0 ) b^x\) [ alternately, \(y = y_0 b^x\) ] The quantity \(f ( 0 )\), or \(y_0\) (pronounced '\(y\) nought' or '\(y\) sub zero') is called the **initial value** of the function.

**Whenever the rate of change of a quantity is proportional to the size of the quantity, an exponential function is involved.**

Consider the charging of compound interest on a loan.

The size of the amount owed increases with the compounding of interest. But interest is a percentage of the amount owed. So the compounding of interest causes the amount owed to **increase** at a rate **proportional** to the amount owed.

Let \(P\) denote the principal amount borrowed. Let \(k\) denote the number of compounding intervals and let \(r\) denote the interest rate per compounding interval (exprerssed as a decimal fraction). Then, if \(A\) is the amount owed, \(A = P ( 1 + r )^k\).

The base of this exponential function is \( 1 + r \) and the initial value is \(P\).

Suppose one borrows $100 from a pawn broker and pays 25% compound interest per month. How much will be the amount of one’s debt after four months? [Remember to represent the 25% interest as \(0.25\).]

See SolutionIt is customary to state interest rates in terms of annual rates. Interest may be compounded annually, quarterly, monthly or daily.

Let \(n\) denote the number of compounding intervals per year and let \(t\) denote the length of time in years before the loan is repaid. Then the amount owed can be computed by the formula

\[A = P \left( 1 + \frac{r}{n} \right)^{nt}\]

In this formula, the base is \(\left( 1 + \frac{r}{n} \right)^n\).

For example, suppose the annual interest rate is 12% compounded quarterly. Then \(r = 0.12\), and \(n = 4\). So the base is \(\left( 1 + \frac{0.12}{4} \right)^4=( 1.03 )^4 = 1.1255\). Thus 12% compounded quarterly for four quarters would be equivalent to 12.55% simple interest for one year.

Compute the base of the exponential function in the example above if interest is compounded **monthly**. **Daily**. What would be the equivalent amount of simple interest in each case if the loan is repaid after one year?

If $1000 is borrowed at 12% interest compounded monthly for three years. What is the amount owed at the end of three years?

See SolutionRadioactive substances decay by changing into a non-radioactive substance. They characteristically do so at a rate proportional to the amount of radioactive substance present at any given time. After an interval of time \(H\) called the **half-life** of the radioactive substance, only half the initial amount will still be radioactive. The amount of radioactive substance present at any given time is an exponential function of time.

Let \(R\) denote the amount of a radioactive substance present at time \(t\) and let \(R_0\) denote the amount present initially. Then there is some base \(b\) such that \(R = R_0 b^t\).

When \(t = H\), only half the substance will be radioactive, so \(R = 0.5 R_0\) when \(t=H\) . Thus, \(0.5 R_0 = R_0 b^H\). So \(b^H = 0.5\). Thus \(b = ( 0.5 )^{1 / H}\). So, \[R = R_0 ( 0.5 )^{ t / H}\]A certain radioactive substance has a half-life of \(H = 10\) years. If \(10\) grams are initially radioactive, how many grams will be radioactive after six years?

See Solution