Solution for Explore! on Page 98

The secant lines approach the tangent line, Y = 2, at the point (0,2) of the curve.

What happens to the graph if all the values in List 1 (L1) are made positive?

The graphing calculator constructs the line Y = 4x - 4 tangent to the curve at (2,4). This tangent line intercepts the y-axis at y = -4.

The actual value of this derivative at x = 1 is y. = 1/2. See Example 1.5. However, the numerical derivative, accessed through the dy/dx option of the CALC (2^nd TRACE) menu, obtains the displayed value.

The numerical derivative, nDeriv( ), (accessed as option 8 via the MATH key) uses the symmetric difference quotient formula,

f. (x) =f(x + e ) - f(x - e )

2e

If the dy/dx option is used, the default for the increment e is .001,. Hence, the source of the discrepancy. However, nDeriv(Y1, x, 1, .00001) yields the precise value .5.

Due to the symmetric difference quotient method for computing dy/dx, the numerical derivative gives the erroneous answer of zero for the slope of f(x) = abs(x) at x = 0.

The point (0,0) is too sharp a point on the curve to possess a well-defined tangent line. Hence, the derivative of f(x) = abs(x) does not exist at x = 0. The numerical derivative must be used with caution at such cusp or other unusal points. Try computing the nDeriv of y = 1/x at x = 0 and explaining how such a result could occur numerically.

At x = 1, the slope of the function f(x) has the value 4, which is identical to the y-value for Y2 at x = 1. Of course, nDeriv(Y1 , x, x) shows the equation 2x + 2.

Where Y2 = f . (x) = 0 (or is negligibly small), you should notice that f(x) has a zero slope, characterized by a minimum or maximum point on the particular function. Of course, f. (x) = 3x² -3, which assumes zero value at x = -1 and 1.

Example 2.7 shows that N. (8) = 21,yielding a 10% rate of change per year. Storing M(t) into Y2, we obtain M. (8) = dy/dx = 18, also yielding a percentage rate of change of 10%. (100M. (8)/M(8) = 100*18/180 = 10% per year.)