MATH 180 FALL 1995 10:00 SECTION FONG

ASSIGNMENT #13

2.6, #4. The tangent line to y = f(x) at x = 0.6 is the line through (0.6, 3.9) with slope f′(0.6). An approximate value for f′(0.6) is given by the average of the average rates of change on the intervals [0.4, 0.6] and [0.6, 0.8], that is, the average rate of change on the larger interval [0.4, 0.8]:

f′(0.6) ≈ \frac{f(0.8) − f(0.4)}{0.8 − 0.4} = 1.25

The equation of the tangent line is then y − 3.9 = 1.25(x − 0.6) or y = 1.25x + 3.15. Using the tangent line to estimate f(x) near 0.6, we find that

f(0.7) ≈ 4.025, f(1.2) ≈ 4.65, f(1.4) ≈ 4.9.

0.7 is much closer to 0.6 than 1.2 or 1.4 in the scale of the problem, so one might “feel most confident about” the estimate for f(0.7).

2.6, #5. x = 7.0 and x = 8.5 are inputs in the table, so we approximate f′(7.0) and f(8.5) by the average of the average rates of changes on two adjacent intervals as in #4:

\begin{aligned}f′(7.0) & ≈ \frac{f(7.5) − f(6.5)}{7.5 − 6.5} = 6.5 − 10.3 = −3.8, \\ f′(8.5) & ≈ \frac{f(9.0) − f(8.0)}{9.0 − 8.0} = 3.2 − 5.2 = −2. \end{aligned}

x = 6.75 is between the inputs 6.5 and 7.0, so we approximate f′(6.75) by the average rate of change on [6.5, 7.0]:

f′(6.75) ≈ \frac{f(7.0) − f(6.5)}{7.0 − 6.5} = \frac{8.2 − 10.3}{0.5} = −4.2.

Since f′(7) = −3.8, the tangent line at x = 7 is approximately y − 8.2 = (−3.8)(x − 7.0) or y = −3.8x + 34.8. In particular, the tangent line gives f(6.8) ≈ (−3.8)(6.8) + 34.8 = 8.96.