MATH 180 FALL 1995 10:00 SECTION FONG

ASSIGNMENT #10

2.3, #1. The slope at every point of the graph is constant, approximately −2, so the derivative function is the constant function y = −2 on the interval [−1, 3].

2.3, #4. The given curve flattens out at x = −2.5 and x = 0 so the derivative is zero at these points. Its slope is positive on the intervals [−4, −2.5] and [0, 4], so the derivative is positive on these intervals with magnitude corresponding to the steepness of the slope. Similarly, the derivative is negative on [−2.5, 0] with magnitude corresponding to the steepness of the slope.

2.3, #6. The given curve flattens out at x = −2 and x = 1 so the derivative is zero at these points. Its slope is positive on the interval [−3.7, −2]. As we move from left to right in this interval the slope starts out steep, becomes gradually less steep, and then flattens out at x = −2. So the derivative on [−3.7, −2] is a decreasing positive function with a plunge to zero at x = −2. The slope of the given curve is negative or zero on [−2, 3.3]. As we move from left to right in this interval, the slope starts at zero, becomes negative but rises to zero at x = 1 and then becomes increasingly more negative. So the derivative on [−2, 3.3] is a non-positive function with zeros at x = −2 and x = 1, decreasing rapidly after x = 1.

2.3, #8. The curves are respectively (a) increasing and concave-up, (b) increasing and concave-down, (c) decreasing and concave-up, (d) decreasing and concave-down.

2.3, #18. f(x_{3}) is the greatest since it is the highest of the given points, f(x_{4}) is the least since it is the lowest of the given points. The steepest positive slope is f′(x_{5}); the steepest negative slope is f′(x_{3}).