Expert: Paul Klarreich - 11/29/2005

Question
Suppose that f is continuous on [-3,3] and differentiable on (-3,3)with f'(x)>=-5 for each x in (-3,3). If f(-3)>=-1, show that f(3)>=-31

Answer
You sent this earlier.  Use these principles:
The concepts here are the interpretations of the first derivative:

(A) If f'(x) is positive on an interval, then f(x) must be increasing on that interval.

(B) If f'(x) is negative on an interval, then f(x) must be decreasing on that interval.

(A1) If f'(x) >= m  on some interval, then the graph is rising MORE STEEPLY than a straight line graph with slope equal to m.

(B1) If f'(x) <= -m  on some interval, then the graph is falling MORE STEEPLY than a straight line graph with slope equal to -m.

(A2,B2) would be similar, for graphs that are LESS STEEP.

Here's what you do:
1. Find the equation of the LINE whose slope is = -5, passing through the first point, (-3,-1).
2. Show that this line would pass through (3,-31)
3. Argue that f'(x) is falling more steeply than this line, therefore, f(3) must be lower than -31.