MATH 180 FALL 1995

ASSIGNMENT #12

2.5, #1. (a) If f′′ is positive on an interval, then f′ is increasing on that interval, and f is concave-up on that interval. (b) If f′′ is negative on an interval, then f′ is decreasing on that interval, and f is concave-down on that interval.

2.5, #2. The right half of a U-shaped curve has first and second derivatives which are everywhere positive; the left half of the same curve has a first derivative which is everywhere negative and a second derivative which is everywhere positive. The right half of a curve in the shape of an upside-down U has first and second derivatives which are everywhere negative; the left half of the curve has a first derivative which is everywhere positive and a second derivative which is everywhere negative.

2.5, #3. For (a) we can join two curves from #2, beginning with right half of a U followed by the left half of an upside-down U. (b) The graph of the first derivative is positive everywhere with the shape of an upside-down U. (c) The graph of the second derivative is a decreasing function whose first half is positive and whose second half is negative. So the graph of the second derivative could resemble a line of negative slope.

2.5, #8. We approximate f′(0.6) by the average of the average rates of change of f on the intervals [0.4, 0.6] and [0.6, 0.8] since this uses the data on both sides of 0.6. This gives f′(0.6) ≈ 1.25. We approximate f′(0.5) by the average rate of change of f on [0.4, 0.6] since [0.4, 0.6] is the smallest interval in the data containing 0.5. This gives f′(0.5) ≈ 2. Since we have values for f′(0.6) and f′(0.5), we may approximate f′′(0.6) by (f′(0.6) − f′(0.5))/(0.6 − 0.5). This gives f′′(0.6) ≈ −7.5. The maximum and minimum values of f occur near x = 0.8 and somewhere between 0.2 and 0.4.

Comments: A number of students found #3 difficult, particularly the part asking for a sketch of f′′(x). The relation of f′′(x) to f′(x) is the same as the relation of f′(x) to f(x). So if the first part of the problem gave no difficulties, neither should the second part.