SOLVED Show that sqrt(1+x) <1+(1/2)x if x>0 ??

Jul 2010
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I have another test question i got wrong and didnt undertand soo im hoping someone can help me out.

Show that sqrt(1+x) <1+(1/2)x if x>0 or ( formally written in attachment)
MSP160019bb47e422a3f76500004b16f392eaba4c0f.gif

I used the formula f'(c)= [f(b)-f(a)]/(b-a) and then just ended up with a mess at the end which is probably why i got it wrong. How would i go abouts starting this problem?
 
Last edited:
Jun 2008
125
14
Plymouth
Just square both sides

\(\displaystyle 4+4x<4+x^2+4x\)

\(\displaystyle x^2>0\)
 
Jun 2008
125
14
Plymouth
Multiply both side by 2. \(\displaystyle 2\sqrt{1+x}<x+2\)
 
Dec 2009
3,120
1,342
I have another test question i got wrong and didnt undertand soo im hoping someone can help me out.

Show that sqrt(1+x) <1+(1/2)x if x>0 or ( formally written in attachment)
View attachment 18197

I used the formula f'(c)= [f(b)-f(a)]/(b-a) and then just ended up with a mess at the end which is probably why i got it wrong. How would i go abouts starting this problem?
If x is positive, prove

\(\displaystyle \sqrt{1+x}<1+\frac{1}{2}x\)

\(\displaystyle 1+\frac{1}{2}x=\frac{2+x}{2}\)

which is the positive square root of \(\displaystyle \frac{4+4x+x^2}{4}\)

Taking it for granted that we are referring to the positive square root of 1+x, then

\(\displaystyle \sqrt{1+x}<\sqrt{\frac{x^2+4x+4}{4}}\ ?\)

\(\displaystyle \sqrt{\frac{4x+4}{4}}<\sqrt{\frac{x^2+4x+4}{4}}\ ?\)

Yes
 
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Nov 2009
485
184
You can also use calculus to prove the inequality.

Consider the function \(\displaystyle g(x)=\sqrt{x}-(1+\frac{1}{2}x)\). It suffices to show that when \(\displaystyle x>0\), we have \(\displaystyle g(x)>0\). First check \(\displaystyle g(0)=0\) and find \(\displaystyle g'(x)\) and show that it is always positive. Then you can conclude.
 
Dec 2009
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You'd need a variation on that.
If you prove a derivative is always positive,
then you are proving the function is always increasing.

Instead....

\(\displaystyle g(x)=(1+x)^{0.5}-1-0.5x\)

Locate the maxmum value of g(x)

If \(\displaystyle g'(x)=0\ for\ x=k\)

then if g(k)<0 the inequality is proven.



Alternatively

\(\displaystyle h(x)=1+0.5x-(1+x)^{0.5}\)

Find the value of x=k giving minimum h(x),
then show h(k)>0

\(\displaystyle 0.5-0.5(1+x)^{-0.5}=0\)

\(\displaystyle \frac{1}{\sqrt{1+x}}=1\)

x=0.

2nd derivative test at x=0.....\(\displaystyle 0.25(1+x)^{-1.5}\)

this is 0.25 at x=0, indicating a minimum on the curve.

h(0)=0

This means that if x is positive, h(x)>0
 
Nov 2009
485
184
You'd need a variation on that.
If you prove a derivative is always positive,
then you are proving the function is always increasing.
We also require \(\displaystyle g(0)=0\), which I mentioned in my post.

For a more formal conclusion: we have \(\displaystyle g(0)=0\) and \(\displaystyle g(x)>0\) for all \(\displaystyle x>0\). Then, by the Mean Value Theorem, there is \(\displaystyle c\in (0, x)\) such that \(\displaystyle \frac{g(x)-g(0)}{x-0}=g'(c)\). But this simplifies to \(\displaystyle \frac{g(x)}{x}=g'(c)>0\). Multiplying \(\displaystyle x\) to both sides gives the desired result.
 
Jul 2010
86
0
We also require \(\displaystyle g(0)=0\), which I mentioned in my post.

For a more formal conclusion: we have \(\displaystyle g(0)=0\) and \(\displaystyle g(x)>0\) for all \(\displaystyle x>0\). Then, by the Mean Value Theorem, there is \(\displaystyle c\in (0, x)\) such that \(\displaystyle \frac{g(x)-g(0)}{x-0}=g'(c)\). But this simplifies to \(\displaystyle \frac{g(x)}{x}=g'(c)>0\). Multiplying \(\displaystyle x\) to both sides gives the desired result.
Thats the formula i mentioned before but this whole g(x) thing is confusing me
 
Nov 2009
485
184
What part of \(\displaystyle g(x)\) is causing you trouble?