Expert: Paul Klarreich - 5/13/2007

Question
given f(x)=1/x+lnx,defined only on the closed interval
1/e¡Üx<e
a). showing your reasoning,determine the value of x at which f has its i). absolute maximum, ii).absolute minimum
b). for what values of x is the curve concave up?
c).on the coordinate axis provided, sketch the graph of f over the interval 1/e¡Üx<e
d). given that the mean value (average ordinate) of f over the interval is 2/e-1, state in words a geometrical interpretation of this number relative to the graph.
thanks


Answer
Questioner:   tina
Category:  Advanced Math
 
Subject:  calculus
Question:  given f(x)=1/x+lnx,defined only on the closed interval
1/e <= x <= e
a). showing your reasoning,determine the value of x at which f has its i).

absolute maximum, ii).absolute minimum
b). for what values of x is the curve concave up?
c).on the coordinate axis provided, sketch the graph of f over the interval

1/e¡Üx<e
d). given that the mean value (average ordinate) of f over the interval is

2/e-1, state in words a geometrical interpretation of this number relative

to the graph.
thanks
..........................................
Hi, Tina,

I assume you meant to write   1/e <= x <= e.  Don't try to include special

characters in your questions -- they just don't get through the site

correctly.

a) Finding absolute max and min involves:
  1. Find the critical points.  That includes:
       Stationary points:  dy/dx = 0
       Singular points:    dy/dx is undefined.
       Endpoints of the interval.
  2. Test the function values at these, if possible.
  3. Pick biggest and smallest.

f(x) = 1/x + ln x

dy/dx = -1/x^2 + 1/x
 -1 + x
= -------
   x^2
 
 x - 1
= ------
  x^2

Stationary point:  x = 1  (OK, that's in your interval.)
Singular point:    x = 0. (NOT OK.)
Left end point:    x = 1/e
Right end point:   x = e

f(1) = 1 + ln 1 = 1   << smallest
f(1/e) = e + ln 1/e = e - 1, about  1.7   << biggest
f(e) = 1/e + ln e = 1/e + 1, about  1.35
..................................
Concave up:  (I don't like that phrase; I prefer 'turning to the left'.)

f''(x) = 2/x^3 - 1/x^2
 2 - x
= ------
  x^3
Inflection point at  x = 2.  (OK, that's in your interval.)

If  x < 2. then -x > -2, and  2 - x > 0, positive.  Concave UP.
If  x > 2. then -x < -2, and  2 - x < 0, negative.  Concave DOWN.

So you are concave up on  [1/e,2)
.....................
c) sketch.  Sorry, that's going to be hard using only notepad.  
........................
d) If the mean value of a function over an interval is  A, then the areas:

Under f(x) and above  y = A, and
Above f(x) and under  y = A

are equal.