Expert: Paul Klarreich - 5/3/2007

Question
Let f be a function that is even and continuous on the closed interval [-3,3]. The function f and its derivatives have the properties indicated in the table below.
x------0----0<x<1----1---1<x<2--2----2<x<3
f(x)---1----pos------0---neg---(-1)--neg
-f'x--und.--neg------0---neg---und---pos
f''x--und.--pos------0---neg---und---neg
a) find the x-coordinate of each point at which f attains an absolute maximum value or an absolute minimum value. For each x-coordinate you give, state whether f attains an absolute maximum or an absolute minimum.
b) find the x-coordinate of each point of inflection on the graph of f. Justify your answer.
c) in the xy- plane provided below, sketch the graph of a function with all the given characteristics of f.

Answer
Questioner:   h
Category:  Calculus
Private:  No
 
Subject:  calculus
Question:  Let f be a function that is even and continuous on the closed interval [-3,3]. The

function f and its derivatives have the properties indicated in the table below.
x------0----0<x<1----1---1<x<2--2----2<x<3
f(x)---1----pos------0---neg---(-1)--neg
f'x---und.--neg------0---neg---und---pos
f''x--und.--pos------0---neg---und---neg
a) find the x-coordinate of each point at which f attains an absolute maximum value or an absolute

minimum value. For each x-coordinate you give, state whether f attains an absolute maximum or an

absolute minimum.
b) find the x-coordinate of each point of inflection on the graph of f. Justify your answer.
c) in the xy- plane provided below, sketch the graph of a function with all the given

characteristics of f.
.......................................
Hi, h,

[but this is the last time -- in the future please provide a NAME, not a letter.]

You have definite points plotted:  (0,1), (1,0), (2,-1)

Here are the conclusions you will draw:

At x = 1:  f'(x) is negative to the left and again to the right;
  Both f'(1) and f''(1) = 0, so this is an INFLECTION POINT.
  Since f(1) = 0, your inflection point is (1,0)

At x = 2:  f'(x) is negative to the left and positive to the right;
  So the graph is falling down to (2,-1) and then rising, but there is no horizontal tangent.

(0,1) is an ABSOLUTE MAXIMUM, because f(x) decreases from that point to (1,0), continues to decrease to (2,-1).  It increases from (2,-1) to x=3, but REMAINS NEGATIVE, so it can't be higher than f(x) = 1.

(2,-1) is an ABSOLUTE MINIMUM, because f(x) has been decreasing for every value to the left, and increasing for every value to the right.

(1,0) is your point of inflection, since  f''(x) is negative to the left and positive to the right. [A point of inflection separates f''<0 from f''>0.]

For your graph, draw your curve starting at (0,1), going down to (1,0), going down again to (2,-1), then going up, but not crossing the x-axis.

When you get to (1,0) be careful.  The graph must come down smoothly and be just about horizontal, then be just about horizontal on the right, too, as it continues downward.