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July 14th 2010, 12:06 PM
kensington

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

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)
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?
July 14th 2010, 12:12 PM
fobos3

Just square both sides

$4+4x<4+x^2+4x$

$x^2>0$
July 14th 2010, 12:24 PM
kensington

Quote:

Originally Posted by fobos3

Just square both sides

$4+4x<4+x^2+4x$

$x^2>0$

Im a tad confused where the 4's came from.
July 14th 2010, 12:50 PM
fobos3

Multiply both side by 2. $2\sqrt{1+x}<x+2$
July 14th 2010, 01:32 PM
Archie Meade

Quote:

Originally Posted by kensington

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)
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

$\sqrt{1+x}<1+\frac{1}{2}x$

$1+\frac{1}{2}x=\frac{2+x}{2}$

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

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

$\sqrt{1+x}<\sqrt{\frac{x^2+4x+4}{4}}\ ?$

$\sqrt{\frac{4x+4}{4}}<\sqrt{\frac{x^2+4x+4}{4}}\ ?$

Yes
July 14th 2010, 02:00 PM
roninpro

You can also use calculus to prove the inequality.

Consider the function $g(x)=\sqrt{x}-(1+\frac{1}{2}x)$ . It suffices to show that when $x>0$ , we have $g(x)>0$ . First check $g(0)=0$ and find $g'(x)$ and show that it is always positive. Then you can conclude.
July 14th 2010, 02:45 PM
Archie Meade

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....

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

Locate the maxmum value of g(x)

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

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

Alternatively

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

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

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

$\frac{1}{\sqrt{1+x}}=1$

x=0.

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

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
July 14th 2010, 05:07 PM
roninpro

Quote:

Originally Posted by Archie Meade

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 $g(0)=0$ , which I mentioned in my post.

For a more formal conclusion: we have $g(0)=0$ and $g(x)>0$ for all $x>0$ . Then, by the Mean Value Theorem, there is $c\in (0, x)$ such that $\frac{g(x)-g(0)}{x-0}=g'(c)$ . But this simplifies to $\frac{g(x)}{x}=g'(c)>0$ . Multiplying $x$ to both sides gives the desired result.
July 14th 2010, 05:15 PM
kensington

Quote:

Originally Posted by roninpro

We also require $g(0)=0$ , which I mentioned in my post.

For a more formal conclusion: we have $g(0)=0$ and $g(x)>0$ for all $x>0$ . Then, by the Mean Value Theorem, there is $c\in (0, x)$ such that $\frac{g(x)-g(0)}{x-0}=g'(c)$ . But this simplifies to $\frac{g(x)}{x}=g'(c)>0$ . Multiplying $x$ to both sides gives the desired result.

Thats the formula i mentioned before but this whole g(x) thing is confusing me
July 14th 2010, 05:16 PM
roninpro

What part of $g(x)$ is causing you trouble?
July 14th 2010, 05:42 PM
kensington

Quote:

Originally Posted by roninpro

What part of $g(x)$ is causing you trouble?

So im looking at my original question with the square roots and inequality and i dont see how i can take those values and sub then in the mean value Theorem.
I basically wote it as [1+(1/2)x - sqrt(1+x)]/ [(1/2)-(1+x)^(1/2)]... then eaither i got that wrong or i messed something up in simplifing because i couldnt prove thats its >0
July 15th 2010, 05:59 AM
Archie Meade

1 Attachment(s)

Do you want to analyse that with the mean value theorem?

An algebraic or graphical solution is quite basic..

$f(x)=\sqrt{1+x}$

$g(x)=\frac{2+x}{2}$

$\sqrt{1+x}=\sqrt{\frac{4+4x}{4}}$

$\sqrt{\frac{4x+4}{4}}=\sqrt{\frac{x^2+4x+4}{4}}$

only when x=0.

Hence if x is positive, and since both functions are increasing functions

$f'(x)=\frac{1}{2\sqrt{1+x}}$

$g'(x)=\frac{1}{2}$

For x>0, g'(x)>f'(x) which means that after x=0, g(x) is increasing more than f(x)
for all x>0.

Attachment 18220