Solution for Explore! on Page 197

The functional setup and graphs appear below. Use the TRACE feature to locate where f ‘(x) intercepts the x-axis (the y-value of 2E^-6 = .000002 can be consider neglible).

You can observe that the graph of f(x) in bold is decreasing between x = -2 and x = 1, the same interval over which f ‘(x) is negative. (Note that its graph is below the x-axis.) Where f(x) is increasing, f ‘(x) is positive. It appears that where the derivative is zero, namely, at x = -2 and 1, the graph of f(x) has turns or bends in the curve, indicating high or low points.

A high point on the left part of the graph before the asymptotes occurs at the origin for both curves, but the low point on the right branches of either curve shifts from x = 4 to x = 8. Note that these low points actually have a larger y-value than any point on the left branch of the curves. The intervals over which g(x) decreases is [0,4) and (4, 8].

Evaluating the graphs, there appears to be no change to the x coordinates of the critical points, although the y-values have obviously been increased by 10. There is a definite effect on the roots of the functions, with f(x) having two real roots while g(x) having none. At x = -3 both graphs have a critical value that is neither a relative minimum or relative maximum. It appears similar to a skier’s mogul, with a momentary zero slope at that point on a downhill run.

It can be verified that the relative extrema of f(x) are the zeroes of f ‘(x), for the values shown in the graphs. What are these values in radical form?

Over the interval [-2, 1], the point (-.865, 6.065) becomes the absolute maximum of f(x), but the point (1,0) is its absolute minimum in that interval. The relative minimum at x = 1.535 is outside the given interval.

Because f "(x) = 6x -2, f "(2) = 10, a positive value. This value indicates a positive curvature for f(x) at (2, 0). The second derivative of f(x) is zero at x = 1/3, the point at which f(x) appears to change concavity from downward to upward. This x-coordinate is also the relative (absolute) minimum of the slope function f ‘(x).

The graphing setup appears below and the pairs of graphs are consecutively presented. The critical values of f(x) are also the zeroes of f ‘(x), namely x = -1.5 and 1, the latter a double root.

The second derivative f"(x) shows that there are two points of inflection, when x =

2/3 and when x = 1. The root finding feature of your graphing calculator can find the zeroes of f"(x).

Graphically it appears that lim f(x) = - ¥ , by tracing f(x) close to the left of 4.

A table display can also be used. Just set TblStart = 3.94 and D Tbl = .01.

To demonstrate lim f(x) = 1 as x approaches ¥ , observe how close f(x) approaches the vertical asymptote y = 1. You can observe this by tracing out large values of x. Algebraically f(x) can be written as 1 + 2/(x-4) and for large x values the 2/(x-4) term becomes negligible.

The graph using the standard window shows vertical lines at x = -5 and x = 3 that visually appear as vertical asymptotes. These lines are actually attempts by the calculator to connect points across the asymptote. Note that he decimal variant window does not display these lines.

As you trace the graph for large x-values, f(x) hovers below but approaches the horizontal line, y = 3, visually demonstrating that this is a horizontal asymptote. Can you find a relative minimum on this right branch of f(x)? Also how far out on the x-axis do you need to trace to obtain a y-value above 2.9?

The function f(x) is not defined for x £ 2, in particular at x = 2. It has no absolute maximum in the interval [1,3], although a relative maximum occurs at x = 3. An absolute minimum occurs at x = 2.4, but the calculator may give a very close approximation.

NOTE: The extreme value property does not apply to the example above because the interval in question is actually (2, 3], half open-half closed.

Using x in place of t to graph S₁ (t) and S(t), set the maximum value of 46 mph into Y₃. Using the intersection feature of the graphing calculator we find that this speed occurs twice a day using S₁ (t), at t = 1.0428 (1:03 pm) and t = 3.2381 (3:14 pm).