Expert: Paul Klarreich - 11/23/2005

Question
               1.  a.  If f(1)=10 and f’(x) >= 2 for 1 <= x <= 4 , then how small can f(4) possibly be ?
      b.     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
            2.  Find the  maximum and minimum values of the functions on the given intervals.
a. f(x)= -3x^2/3 , [1,1]                          
b. g(x) = (x5 + x + 1)^-1 , [-1/2 , 1]

1.   A particle moves along the x-axis , its position at a time t is given by
X(t)= t / 1+t^2  , t >= 0 ,where t is measured in seconds and x in meters.

a.   Find the velocity at any time t .
b.   Find the acceleration at any time t , when it is 0?
c.   Find the interval of time when it is advancing and when it is not ?


Answer
Dessalegn,

Your question came in somewhat garbled.  I have had to guess what you meant in some places.  For example:

a. If f(1)=10 and f’(x)

Is that  f'(x)??

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.

 1. a. If f(1)=10 and f'(x) >= 2 for 1 <= x <= 4 , then how small can f(4) possibly be ?

The graph of f(x) passes through (1,10).  A stright line with slope equal to 2 that passes through (1,10) would pass through (4,16).  Use (A1): This graph rises more steeply than that line, so f(4) must be >= 16.

   b.    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

Got the idea?  Try Principle B2 on this.  The graph passes through (-3,-1) and although it is falling, it falls LESS STEEPLY than a line with slope m = -5.  

You could derive the equation of that line:
y - (-1) = -5(x - (-3)  [look at all those minuses!]
y + 1 = -5x - 15
y = -5x - 16

Now put x = 3
y = -15 - 16 = -31

So the line would drop to y = -31, and the graph drops LESS STEEPLY.  So the function at +3 must >= -31.


2. Find the maximum and minimum values of the functions on the given intervals.
a. f(x)= -3x^2/3 , [1,1]    (You mean [-1,1], don't you?)               
b. g(x) = (x5 + x + 1)^-1 , [-1/2 , 1]

Basic Max-min theory:  If a function is cont and diff on [a,b], then the max or min must occur at one of these points:

  A. Where f'(x) = 0
  B. Where f'(x) is undefined.
  C. At an endpoint.

a. f'(x) = -2x^(-1/3) = -2/(cube root of x)
This is undefined at x = 0.  So the rest is easy:  The candidates for max-min are x = 0, x = -1, x = 1.  Just plug those into the ORIGINAL FUNCTION and pick your max and min.
            -(4x^4 + 1)
b. g'(x) = --------------
          (x^5 + x + 1)2

Now we look for points where the numerator is zero.  There aren't any.  x^4 is always non-negative, so 4x^4 + 1 is never zero.  No sense looking at the denominator, either.  If it is zero, then the original function is undefined, too.

So you are left with the endpoints as your candidates.  Go to it.


1. A particle moves along the x-axis , its position at a time t is given by
X(t)= t / 1+t^2 , t >= 0 ,where t is measured in seconds and x in meters.

You mean:  X(t)= t /(1+t^2)


Here, the principles look different, but really aren't:

Writing  v(t) = X'(t) as the velocity, then a(t) = v'(t) = X''(t):
---------------------------------------------



(A) v > 0  means moving to the right. (advancing?)
(B) v < 0  means moving to the left.
(C) v = 0  means stationary.


a. Find the velocity at any time t .

Use the quotient rule:
      (1 + t^2)(1) - (t)(2t)   1 + t^2 - 2t^2    1 - t^2
v(t) = ---------------------- = --------------- = --------
          (1 + t^2)^2             ()^2             ()^2



b. Find the acceleration at any time t , when it is 0?

For a(t), just differentiate that again, using the quotient rule.  I'll leave that to you.  It's a bit messy, but....

And when you are done, just set the NUMERATOR equal to zero and solve the equation.


c. Find the interval of time when it is advancing and when it is not ?

You want to know when is  v(t) > 0?

Set  1 - t^2 > 0,   1 > t^2,  or t^2 < 1

Take square root of both sides, so t < 1.  (Since t >= 0, we don't have to worry about negative values.)

Conclusion: Advancing when t < 1

Not advancing when t >= 1