# 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.)

Given

Show that

Note: Don't try to find

##### 2 Answers

We have:

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

By the Mean Value Theorem,

# 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) = { (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

# :. (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

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

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

#### Explanation:

We provide an algebraic proof of this below:

and

Thus

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