Show that a g(x) has a unique fixed point using the theorem

[xet]

Question:
Show that g(x)=(x/2)+ (1/x) has a unique fixed point in the interval [1,2] using a fixed point theorem.
Do not approximate the fixed point. Verify all conditions.

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My work so far:
We see on the graph that theres an intersection with g(x) and y=x and its a unique fixed point because a fixed point of g(x) are solutions of g(x)=x.
g(1.4142136)= 1.4142136

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My problem:
Not sure how to prove it, i think ive somewhat have the start, maybe, but have no idea on how to do the other part of the fixed point theorem.
Please Help!!!! thanks in advance

 

[BigGlenntheHeavy]

Use the Intermediate Value Theorem to prove that any continuous function with domain [1,2] and range [1,2] must have a fixed point.

 

[daon]

Why not just solve \(\displaystyle \frac{1}{x}=\frac{x}{2}\) and verify that it belongs to the domain and range?

 

[xet]

How can i prove that the fixed point is unique? im bad at proving stuff,

Written below is how prof used to prove a fixed point is unique but not sure how to make it apply to my exercise.

Assume there are two distinct fixed points, p and q,,, p does not equal q
that is g(p)=p and g(q)=q.
|p-q|=|g(p)-g(q)|= |g'(x)(p-q)| = |g'(x)||p-q| <= k |p-q| < |p-q|
and this is a contradiction
therefore the fixed point is unique.

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please help, thanks in advance

 

[daon]

" <= k |p-q| < |p-q| "... how do you justify this? You are assuming k<1?

Using just simple algebra you get that the fixed point must be one of \(\displaystyle (\sqrt{2},\sqrt{2}), (-\sqrt{2},-\sqrt{2})\), only one of which resides in \(\displaystyle [1,2]x[1,2]\).

Or, more "fancily", if p and q are not equal then g(p)-g(q)=p-q => (p-q)/2+(q-p)/qp = p-q => (p-q)qp=2(q-p) => qp=-2, which means one of p or q must be negative, a contradiction (on our domain).