Solution for Explore! on Page 300

One method to display all the desired graphs is to list all the desired bases in the function form. One can observe that for b > 1, the functions increase exponentially, all passing through the point (0, 1). Why? Note that y = 4^x will lie between y = 2^x and y = 6^x. Likewise, the graph of y = e^x would lie between y = 2^x and y = 3^x.

C: First, y = (-1)^x is only defined for certain rational values of x. Try y = (-1)^{n / m} for different n and m integers in your calculator. Do you perceive a pattern? For example, for x = .2 = 1/5, y = -1, but for x = .3 = 3/10, y is undefined. In the graph below (using a decimal window) y is plotted for decimal values of x, and dots appear at +1 and -1 for only certain decimal values of x. Next the graphs y = b^x, for b = .5, .25 and .1 also appear below. These simply the reciprocals of the graphs of y for b = 2, 4 and 10. For example, y = .25^x = 1/(4^x).

The graph appears to have a vertical asymptote between y = 2 and 3. The table value for multiples of x = 100,000 shows a limit value of about 2.71828. Large negative x-values show the same limit. This irrational number is given the symbol "e."

Using the STO® key, store the value 10 to the variable T, .05 to R, and 12 to K. Then evaluate the expression. Change the values of R to .06 and K to 4 and use the ENTRY feature (2^nd ENTER) to retrieve the expression to re-evaluate. The 6% quarterly compounding produces a much better investment.

The 6% quarterly compounding has already been completed in the previous exercise. The monthly and continuous compounding appear below. The continuous compounding yields the largest interest. For T = 15 years, monthly and continuous compounding are also shown.

Place an initial value into P, say, 1000, and then press SOLVE (ALPHA ENTRY) to obtain the present value for P. You can access the Equation Solver through the MATH menu by selection option 0.

For Example 1.2, the following screens show first the discrete compounding solution and then the continuous compounding solution.

The equation solver setups are given below. The number of bacteria is essentially 54,000.

The equation solver setups are given below. The original value V changes to $18,096.75.

The DRAW (2nd PRGM) key has an option to draw an inverse relation, 8:DrawInv. Apply this option to the function Y1. Note that the graph of y = ln x is symmetric to that of y = e^x about the line y = x. This implies that e^lnx = x = ln(e^x). (The graph of y = ln x cannot be traced since it is only a drawing.)

There appear to be two solutions, x = -4.339 and x = .8849. You should numerically verify that these two values are solutions of the given equation.

It appears that the graph of y = log(x), and not y = ln(x) is the symmetric reflection of graph of y = x, so that these two functions are inverses of each other. Thus, log(10^x) = 10^logx = x. For example, 10^log5 = 5.

Place the equation 0 = F - P*e^(R*T) into the equation solver and set F = 5,000, P = 2,500 and R = .085. Solving for T yields a doubling time of 8.15 years.

At the city center, x = 0 and Q = 15,000. Hence, A = 15,000. Next use the equation solver to determine that when x = 10 and Q = 9,000, K = .051 approximately. Finally, set Q = 13,500 and solve for x, obtaining the value 2.06 miles from city center.

For x = 4, the equation of the tangent line is given below. The slope of this line is approximately .25, which is the reciprocal of 4. The numerical discrepancy occurs because the derivative is calculated using the numerical derivative with a tolerance level of .001.

For x = 7, the y-coordinate and the derivative value at this point coincide to the fifth decimal value. Convince yourself that at other values of x, the derivative of e^x is identical to itself.

When A = 1 and B = 1, the maximal y-value occurs at x = 1 (first figure). When A = 1 and B = .1, it occurs at x = 10 (2nd figure). When A = 10 and B = 0.1 this maximum remains at x = 10 (3rd figure). In this case the y-coordinate of the maximum increases by the A factor.

In general, it can be shown that the x-coordinate of the maximum point is just 1/B. The A factor does not change the location of this x-coordinate, although it affects the y-value as a multiplier. To verify this analytically, set the derivative of y = Axe^-Bx and solve for the location of the maximal point.

The functional setup in the equation editor is shown below, yielding the displayed graphs. Using the trace function, you can determine that two x-intercepts of f "(x) are located at the approximate values of x = 0.6 and x = 3.4. Since the second derivative f "(x) represents the concavity function of f(x), we know that at these x-values f(x) changes concavity from positive (concave upwards) to negative (concave downwards) at x = 0.5858 and from negative to positive at x = 3.4142. Thus, these are appropriately labeled inflection points.

Using the minimum finding feature of the graphing calculator you will obtain what appears to be the point (2, -1.545). Note that the x-value is approximately 2 on the calculator display. The x-intercept for f ‘(x) confirms the value x = 2.

A quick view of table values for Y1 in increments of 10 years shows that a maximum in the range of the 70,000 occurs between 40 and 60 years. Hence, a reasonable window for viewing the salient part of the graph is [40, 60]10 by [600,000, 800,000]100,000. Using the maximum finding feature of the graphing calculator pinpoints the location (51.02, 711,347.35) as the maximum point.

As x approaches 10 (weeks) the function attains a value close to 20 thousand people infected (Y > 19.996 thousand). Setting Y2 = 18 (thousand) and using the intersection feature of the graphing calculator, you can find that 90% of the population becomes infected when x = 4.28 weeks (about 30 days).