1. Let t denote time, measured in hours, starting at noon. The table below gives the temperature (in degrees Fahrenheit), over a 24 hour period, as a function f of time. Use the table to answer the following questions.

a) What is the value of f(14) and state in words what this value represents.

f(14) is "the temperature in degrees Fahrenheit 14 hours hours after noon" or the temp at 2 AM which is -20 deg F.

b) Was it colder at t = 4 or t = 16?

f(4) = 4 and f(16) = -10, it was colder at t = 14.

c) At what approximate time(s) of day was the temperature 3 degrees Fahrenheit?

Somewhere between t = 4 and t = 6, and again between t = 18 and t = 22. Algebraically, from t = 4 to 6, the temperature is changing at about (0 - 4) / (6 - 4) = -2 degrees F per hour. Thus, it will take about 1/2 an hour to drop from 4 to 3 degrees, so 4:30 PM is one estimate. Graphically, the temperatures from t = 16 to 22 are plotted below along with temp = 3. By reading the graph, we see at about t = 20.4 (or 8 AM + 0.4*60 = 8:24 AM, the temperature is approximately 3 degrees.

> plot({[[16,-10],[18,0],[22,5]],[[16,3],[22,3]]},color=black,
labels=[`t`,`temp`]);

d) Was the temperature rising or falling at t = 14?

From t = 12 to 16, the temperature goes from -23 deg F to -10 deg F, so it appears to be rising at t = 14.

e) What is the average rate of change in temperature from noon to midnight?

(-23 - 20) / (12 - 0) = -3.5833 deg F per hour.

f) Estimate (as best you can) the instantaneous rate of change in temperature at 2 pm.

(temp(4PM) - temp(noon)) / (4 - 0) = (4 - 20) / 4 = -4 deg F per hour.

2. A bug starts out 10 feet from a light, flies closer to the light, then farther away, then closer than before, then farther away. Finally the bug hits the bulb and flies off. Sketch a possible graph of the distance of the bug from the light as a function of time.

Your graph should start at (0, 10), decrease, increase (perhaps above y = 10), decrease to a lower y value than before, increase (perhaps to a new maximal distance), then decrease to zero and increase "away", something like:

3. How do you decide if a table of values for a function f(x) can be closely approximated by a linear formula f(x) = m*x + b?

Check all of the "consecutive slopes" y / x, if they are all about the same, a linear formula will fit well.

4. Shown below is a graph of the function f(x), as x runs from 0 to 5. Use the plot to answer the following questions.

a) What is f(2)?

When x = 2, y is about -1.3, so f(2) = -1.3.

b) For what value of x does f(x) = 2?

The graph of f(x) intersects y = 2 at about x = 4.5, so f(4.5) = 2.

c) For which values of x is f(x) decreasing?

From the "hilltop" near x = 0.3 to the "valley bottom" near x = 3.4.

d) Draw in the tangent line to f(x) at x = 2 and then estimate the slope of this tangent line.

Drawing in a line which "kisses" the graph of f(x) at x = 2 (shown in red above), yields a line which goes through the points (1, 2) and (3, -4.7), so the slope of the tangent line is roughly (-4.7 - 2) / (3 - 1) = -3.35.

e) What (approximately) is the instantaneous rate of change of f(x) at x = 2?

By definition, this is the slope of the tangent line to f(x) at x = 2, or approximately -3.35.

f) For which values of x is f(x) < ?

The graph appears to be below the gridline at y = -3 from x = 2.4 to x = 4.1.

g) For x values between 0 and 5, how many solutions does f(x) = 0 have? What are they?

There are 2, one near x = 1.5 and the other near x = 4.4.

h) Sketch a graph of .

Flip the graph about the x-axis, then move it down by 3:

5. Use parametric equations to sketch a graph of P(t) = 5*0.6^t, and its inverse, for values of t from 0 to 5.

P(t) is red, P inverse is blue. The Y = screen (in parametric mode) of a TI-83 is shown as well. Sorry, this is not a good "picture" of the property that y = x is a mirror for these 2 functions.

6. Find (accurate to 3 decimal places) all solutions (values of w) to 8*w^3 = 1.2^w.

The plot below shows the classical shape of the cubic 8*w^3 (red) and the exponential 1.2^w (blue). There won't be any more solutions for negative values of w (the cubic becomes more negative, while the exponential gets closer to the w axis). However, the exponential will definitely catch and pass the cubic (eventually), so there must be another solution for some larger value of w. Using the intersection or root or zero command on your calculator, you should determine that the crossing point shown here is about w = 0.5159258

> plot([8*w^3,1.2^w],w=-5..10,y=-10..10,color=[red,blue]);

> crossing1:=fsolve(8*w^3=1.2^w,w=0..2);

By scrolling farther and farther to the right, you'll eventually find the last crossing near w = 85. Again the calculator will locate this solution as w = 84.387766. From this point on, the exponential will lie above the polynomial.

> plot([8*w^3,1.2^w],w=0..100,color=[red,blue]);

> crossing2:=fsolve(8*w^3=1.2^w,w=80..90);

7. Find possible formulas for each of the functions represented below.

a) P(t) is the size of a population (after t years) which starts at 250 and decreases by 4% each year.

Constant relative decay, starting at P0 and decreasing by r percent per year is given by P0*(1 - r / 100)^t, thus

b) The functions f(x) and g(x) are given by the following table:

Using your calculator, you should be able to check these "slopes". Note Dx = 2.2.

> f_slopes:=(6.25-9.42)/(3.5-1.3),(4.15-6.25)/2.2,
(2.75-4.15)/2.2,(1.82-2.75)/2.2;
g_slopes:=(98.1-15.6)/2.2,(610-98.1)/2.2,
(3820-610)/2.2,(23900-3820)/2.2;

Since these are NOT the same, we check the ratios:

> f_ratios:=6.25/9.42,4.15/6.25,
2.75/4.15,1.82/2.75;
g_ratios:=98.1/15.6,610/98.1,
3820/610.,23900/3820.;

The ratios for both f and g are nearly constant, so both functions can be approximated by exponential formulas. The ratios should be b^2.2, since the x values differ by 2.2:

> f:=x->k*b^x;
`f(x+2.2)`/`f(x)`=simplify(f(x+2.2)/f(x));

Well 11/5 = 2.2, and Maple prefers fractions? We need to solve b^2.2 = 0.663 (i.e. b = 0.663^(1/2.2) for f and b^2.2 = 6.25 for g.

> b_for_f:=0.663^(1/2.2);
b_for_g:=6.25^(1/2.2);

Now plug in x = 1.3, f(x) = 9.42 and g(x) = 15.6, along with numeric values for b, and solve for k (assuming both functions have the form k*b^x).

> k_for_f:=solve(9.42=k*0.8296919910^1.3,k);
k_for_g:=solve(15.6=k*2.300189646^1.3,k);

I'd say f(x) = 12*0.8296^x and g(x) = 5.2824*2.3^x. You can check these results by defining Y1 = f(x), Y2 = g(x), and setting up your table as shown in the TBLSET screen below on the left. The table on the right comes close to reproducing the values in the original table.

c) The functions h(w) (thick) and k(w) (thin) are plotted below.

Clearly h(w) is linear and k(w) appears to be exponential. The points (0, 4) and (5, -7) appear to be on the graph of h(w), so the slope is (-7 - 4) / (5 - 0) = -2.2. Since the y-intercept is 4, we have h(w) = -2.2*w + 4. Assuming that k(w) = c*b^w, we have c = k(0) = 2. The point (4, 5.8) appears to be on the graph of k(w), and k(4) / k(0) = b^4, so 5.8 / 2 = b^4, and we have b = 2.9^(1/4) = 1.3. Thus, k(w) = 2*1.3^w.

8. NOTE: Graphs of k(w) and h(w) are shown just above (in problem 7). Let f(w) be the piecewise function defined by

a) Calculate f(-1), f(0), f(1), f(2), and f(3).

Since -1 < 0, we use the first rule for f(-1), and so f(-1) = 3. The value w = 0, satisfies the second test, so f(0) = k(0) = 2 (from the graph). Likewise, f(1) = k(1) = 2.6, and f(2) = k(2) = 3.5. Finally, 3 > 2, so f(3) = h(3) = -2.5 (from the graph).

b) Estimate , , , and 

If w is close to 3, f(w) will equal h(w), so the first limit is the same as = -2.5 (from the graph). For w < 0, f(w) is always equal to 3 (by the first rule), so the second limit is 3. For w > 0 (and close to zero), f(w) = k(w), so we approach w = 0 from the right, on the graph of k(w), and the third limit is therefore 2. To the right of w = 2, values of f(w) come from h(w). As w approaches 2 from the right (on the graph of h(w) ), y gets close to -0.3, so the last limit is (about) -0.3.

9. Find an x-range, centered at x = 0, so that the graph of f(x) = on your x-range lies between y = -1 and y = -0.8.

Setting Y1 = f(x), Ymin = -1 and Ymax = -0.8, and trying different X windows of the form 0 - d to 0 + d, yields (roughly) the window settings Xmin = -.08, Xmax = 0.08, to keep Y1 from leaving the window through the top or bottom.

10. Estimate each of the following limits: , , .

By either evaluating the functions for values of "x close to a", or plotting, you should get limits close to

> Limit((exp(x)-1)/x,x=0)=limit((exp(x)-1)/x,x=0);
Limit((sqrt(`3*x`)-3)/(x-3),x = 3)=
limit((sqrt(3*x)-3.)/(x-3.),x = 3.);
Limit(sin(3*x)/(7*x),x = 0)=
limit(sin(3.*x)/(7.*x),x = 0.);

11. If f(x) = 12 - 3*2.4^x, find a formula for the inverse of f(x). For what value of x does f(x) = -10? What is ?

Set y = 12 - 3*2.4^x and solve for x. Thus, (y - 12)/(-3) = 2.4^x. Taking the natural log of both sides gives ln([12 - y]/3) = ln(2.4^x) = x*ln(2.4), so x = .

In the above formula, y = f(x), so substituting y = -10, gives x = ln(22/3) / ln(2.4) = 2.2758. Here is a check:

> 12-3*2.4^2.2758;

The last question is exactly the same as asking for an x with f(x) = -10, i.e., x = 2.2758.

12. If f(x) = (2^x)* cos(x - 0.3) + 0.5, find the average rate of change of f(x) from x = 1 to b, where b is each of the values 1.2, 0.95, 1.0003 and .999997. Estimate the equation of the tangent line to f(x) at x = 1, and the instantaneous rate of change of f(x) at x = 1. Finally, on a well selected viewing window, plot both f(x) and its tangent line for x values from 0 to 2.

The average rates of change all fit the pattern (f(b) - f(1) ) / (b - 1). Setting Y1 and Y2 as shown below on the left (note that Y1 is turned off) and setting the table to ASKmode on the independent variable, yields the right screen of average rates of change.

The tangent line should therefore have a slope of about m = -0.228, and go through the point (1, f(1) ) = (1, 2.03). An equation is thus y = -0.228*(x - 1) + 2.03. The instantaneous rate of change of f(x) at x = 1 is defined as the slope of the tangent line, which is roughly -0.228. The plot below shows both f(x) and this approximate tangent line.

> plot([(2^x)*cos(x-0.3)+0.5, -0.228*(x-1)+2.03],
x=0..2,color=[blue,red],title=`f(x) and tangent line`);

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