Show #|f(b)-f(a)|<= |b-a|# ?

Given #f:RR->RR# differentiable in #RR# with #f'(x)=(2x)/(x^2+1)# , #AAx##in##RR#

Show that #|f(b)-f(a)|<= |b-a|#

#a,b##in##RR#

Note: Don't try to find #f# - (no given values for constant.)

2 Answers
Steve M
Apr 30, 2018

We have:

# f'(x) = (2x)/(x^2+1) \ \ AA x in RR #

By the Mean Value Theorem, #EE c in RR # such that:

# f(b)-f(a) = f'(c) (b-a) #

Hence, we have:

# |f(b)-f(a)| = |f'(c) (b-a) | = |f'(c)| \ |(b-a) |#

So we can reduce the problem to that of proving that:

# |f'(c)| le 1 #

Which requires us to consider the maximum value of #f'(x)#, by finding its critical points, using its derivative, f''(x). So differentiating wrt #x#, by applying the quotient rule:

# f''(x) = { (x^2+1)(2) - (2x)(2x) ) / (x^2+1)^2 #
# \ \ \ \ \ \ \ \ \ \ = (2x^2+2-4x^2) / (x^2+1)^2 #
# \ \ \ \ \ \ \ \ \ \ = (2-2x^2) / (x^2+1)^2 #

So, at a critical point of #f'(x)#, we require its derivative to vanish:

# :. (2-2x^2) / (x^2+1)^2 = 0 => x^2-1=0 => x=+-1#

When:

# { (x=-1,=>f'(x) = -1), (x=1,=>f'(x) = 1) :} #

To determine the nature of the critical points, we could consider the second derivative of #f'(x)#, ie #f'''(x)#, but equally we can examine a graph of ten function:
graph{y=(2x)/(x^2+1) [-6.24, 6.25, -3.12, 3.12]}

And we can see that:

# { ((-1,1),=> \ "minimum"), ((1,1),=> \ "maximum") :} #

As such we can conclude that:

The range of # f'(x)# is #-1 le f'(x) le 1 => |f'(x)| le 1#

Then returning to the original problem, we can conclude that

# |f(b)-f(a)| le |(b-a) | \ \ \ # QED

Ananda Dasgupta
Apr 30, 2018

As was explained in Steve M's answer in https://socratic.org/s/aQwiERBL , we can prove the result using the mean value theorem, provided we can prove that #|f^'(x)| <1# #forall x in RR#.

Explanation:

We provide an algebraic proof of this below:

#f^'(x) = (2x)/(x^2+1) implies#

#f^'(x) + 1 = (2x+x^2+1)/(x^2+1) = (1+x)^2/(x^2+1)>=0quad forall x in RR#

and

# 1-f^'(x) = (x^2+1-2x)/(x^2+1) = (x-1)^2/(x^2+1) >= 0quad forall x in RR#

Thus #forall x in RR#, we have

#-1 <= f^'(x) <= 1qquad implies qquad |f^'(x)| <= 1#

The rest of the proof proceeds as in https://socratic.org/s/aQwiERBL

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