Question

Consider f (x) = sqrt(1+x^2).
(a) Show that for all c, |f'(c)| < 1.
(b) Hence use the Mean Value Theorem to show that, when x != y,
|sqrt(1+x^2)-sqrt(1+y^2)| < |x-y|

Answer 1

f(x) = sqrt(1+x^2)

with this equation, notice no matter the value of x, it will always be defined.

the derivative is:

x/sqrt(1+x^2)

same equation, just now we have x in the numerator

something to notice -
x = 0, f(x) = 0.
x = 1, f(x) = 1/sqrt2
x = 2, f(x) = 2/sqrt5
x = 3, f(x) = 3/sqrt10

notice the denominator keeps getting bigger and bigger, while the numerator climbs at a constant rate. for instance:

x = 100, f(x) = 100/sqrt10001

another way to put it, is the denominator will always be larger than the numerator. therefore, it has to be less than 1, as it will always be a fraction.

Answer 2

dont really know how to go about answering the second part.. i don't really remember what the mean value theorem is XD

i assume its an extension of the fact that f'(x) will always be less than 1 though

Answer 3

I believe the second part goes something like this:
f(x) is continuous on the interval [a,b] and differentiable on the interval (a,b) for any a,b.

By the MVT there exists:
\[f'(c)=\frac{ f(b)-f(a) }{ b-a }=\frac{ \sqrt{1+b^2}-\sqrt{1+a^2 } }{ b-a } \]
but we know that:
\[|f'(c)|<1\]
So,
\[|\frac{ \sqrt{1+b^2}-\sqrt{1+a^2 } }{ b-a }|<1\]
\[|\sqrt{1+b^2}-\sqrt{1+a^2 }|<|b-a|\]
Now set b=x and a=y,
\[|\sqrt{1+x^2}-\sqrt{1+y^2 }|<|x-y|\]
x != y, as 0<0 DNE.

Could anyone check what I have done is a mathematically correct argument?
Thanks