Expert: Paul Klarreich - 5/21/2007

Question
The graph of function f consists of three line segments with the points (-2,0) (1,4) (2,1) and (4,-1). Let g(x) equal the integral where x is the upper limit, one is the lower limit, f(t)dt.
a) compute g(4) and g(-2).
  I found g(4) to be 5/2 (let 4 be x and solve the integral by computing the area of the closed interval)
  I found g(-2) to be -6, by the same way.
b)find the instantaneous rate of change of g, with respect to x, at x equals 1.
  I think this is where one would set up the limit as h approaches zero of f(a plus h)-f(a)/h. I dont really know.
c. find the absolute minimum of the value of g on the closed interval [-2,4]. Justify your answer. I know that this is where we search for critical numbers of f', or g". I think. I am not sure.
d. the second derivative of g is not defined at x equals 1 and x equals 2. How many of these values are x coordinates of points of inflection of the graph of g?Justify your answer. I know g'(x) is equal to f(x). so I know x equals one is a point of inflection (cuz slope changes from positive to negative) and x equals two is not. so one is a point of inflection.

Thanks for your help!

Answer
Questioner:   Marcella
Category:  Calculus
 
Subject:  calculus
Question:  The graph of function f consists of three line segments with the points (-2,0) (1,4) (2,1) and (4,-1). Let g(x) equal the integral where x is the upper limit, one is the lower limit, f(t)dt.

a) compute g(4) and g(-2).
 I found g(4) to be 5/2 (let 4 be x and solve the integral by computing the area of the closed interval)
 I found g(-2) to be -6, by the same way.

b)find the instantaneous rate of change of g, with respect to x, at x equals 1.
 I think this is where one would set up the limit as h approaches zero of f(a plus h)-f(a)/h. I don't really know.


c. find the absolute minimum of the value of g on the closed interval [-2,4]. Justify your answer. I know that this is where we search for critical numbers of f', or g". I think. I am not sure.

d. the second derivative of g is not defined at x equals 1 and x equals 2. How many of these values are x coordinates of points of inflection of the graph of g?Justify your answer. I know g'(x) is equal to f(x). so I know x equals one is a point of inflection (cuz slope changes from positive to negative) and x equals two is not. so one is a point of inflection.

Thanks for your help!
...............................
Hi, Marcella,

************ VIEW THIS IN COURIER FONT ********

You are saying:

g(x) =

{x
|   f(t) dt
}1

where f(t) is defined by those three segments.
(-2,0) (1,4) (2,1) and (4,-1)

---------------*-------------------
                                  
-----------------------------------
                                  
-----------------------------------
                                  
-------------------*---------------
                                  
---*---+---+---+---+---+---+---+---
 -2  -1   0   1   2   3   4
---------------------------*--------

g(4) =

{4
|   f(t) dt
}1

which is the sum of the two trapezoids:
{2
|   f(t) dt = 1(4+1)/2 = 5/2
}1
and
{4
|   f(t) dt = 2(1-1)/2 = 0
}2

and that sum is 5/2.

g(-2) =

{-2
|   f(t) dt =
}1
{1
|  - f(t) dt =
}-2
which is the trapezoid (well, it's a triangle, actually) whose area is:
3(0 + 4)/2 = 6.

This is  -6

So far so good.  You're doing fine.

.............................

b)find the instantaneous rate of change of g, with respect to x, at x equals 1.

d  {x
-- |  f(t) dt =  f(x)
dx }1
This is a basic property of integrals.  You prove it using the Mean Value Theorem.

So g'(2) = f(2) = 1

If that doesn't satisfy you, it goes this way:

           g(2 + h) - g(2)
g'(2) = lim ---------------
      h->0       h

     {2+h          {2
     |  f(t) dt -  | f(t) dt
     }1            }1
lim   ----------------------
h->0         h


     {2+h
     |  f(t) dt
     }2
lim   -----------
h->0     h

This is the Average Value of the function f(t) on the interval.  But as  h->0, the 'squeeze theorem' says that this is between  f(2) and f(2+h).  And since  h -> 0,  2+h -> 2.  So the average value, since f(x) is continuous, will be the same as f(2).  And f(2) = 1.
.....................................
c. find the absolute minimum of the value of g on the closed interval [-2,4]. Justify your answer. I know that this is where we search for critical numbers of f', or g". I think. I am not sure.

Yes, that seems a good approach.  The critical numbers occur at:

A. Places where  g' = f = 0.
B. Endpoints, x = -2 and x = 4

f = 0 at x = 3 only (well, also at x = -2, but that is already an endpoint.)

g(-2) = -6, from above.
g(4) = 5/2, from above.

g(3) will have to be computed:

{3
|   f(t) dt
}1

which is the sum of the two trapezoids:

{2
|   f(t) dt = 5/2
}1

{3
|   f(t) dt = 1/2
}2

5/2 + 1/2 = 3 (little maids?) is the total sum.

Looks like a minimum of -6 at  x = -2, maximum of 3 at x = 3.
............................
d. the second derivative of g is not defined at x equals 1 and x equals 2. How many of these values are x coordinates of points of inflection of the graph of g?Justify your answer.

I know g'(x) is equal to f(x). so I know x equals one is a point of inflection (cuz slope changes from positive to negative) and x equals two is not. so one is a point of inflection.
. . . . . . . . . . .
I tend to agree.

g' = f, so  g'' = f'.  Your reasoning will go this way:

A point of inflection separates regions where the graph is concave up, with g'' > 0, from regions where it is concave down, with g'' < 0.

If g'' > 0, then  g' = f is INCREASING.
If g'' < 0, then  g' = f is DECREASING.

f increases on  [-2,1] and decreases thereafter.  So  x = 1 is your one and only inflection point.