## Going Crazy over this Mean Value Theorem Question!!!

Limits, differentiation, related rates, integration, trig integrals, etc.
kittie21
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### Going Crazy over this Mean Value Theorem Question!!!

Okay so the question is asking to show that for 0<x<y, sqrty - sqrtx < (y - x)/(2 sqrt x) (using the mean value theorem)

I think I have some valuable information pertaining to the question...but I am so confused about how to solve it I could just be pulling numbers out of nowhere in a desperate attempt to feel smart!

I have the slope of the tangent for sqrty - sqrt x = -sqrt y / y and the slope of the tangent for (y - x)/(2 sqrt x) = -(1/2y)/ y But not sure what I do with this information! any help would be amazingly appreciated!

Martingale
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### Re: Going Crazy over this Mean Value Theorem Question!!!

Okay so the question is asking to show that for 0<x<y, sqrty - sqrtx < (y - x)/(2 sqrt x) (using the mean value theorem)

I think I have some valuable information pertaining to the question...but I am so confused about how to solve it I could just be pulling numbers out of nowhere in a desperate attempt to feel smart!

I have the slope of the tangent for sqrty - sqrt x = -sqrt y / y and the slope of the tangent for (y - x)/(2 sqrt x) = -(1/2y)/ y But not sure what I do with this information! any help would be amazingly appreciated!
$\frac{f(y)-f(x)}{y-x}=f'(c)$

$c$ between $x$ and $y$

let $f(t)=\sqrt{t}$

...

kittie21
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### Re: Going Crazy over this Mean Value Theorem Question!!!

Thanks for the response! But I am not sure what I am supposed to do with this f(t) = sqrt t ... how does that fit in with the question? Am I using substitution with the t?

Martingale
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### Re: Going Crazy over this Mean Value Theorem Question!!!

Thanks for the response! But I am not sure what I am supposed to do with this f(t) = sqrt t ... how does that fit in with the question? Am I using substitution with the t?

$\frac{f(y)-f(x)}{y-x}=f'(c)$

$c$ between $x$ and $y$

let $f(t)=\sqrt{t}$

...
$f'(t)=\frac{1}{2\sqrt{t}}$
$f'(c)=\frac{1}{2\sqrt{c}}$

$\frac{f(y)-f(x)}{y-x}=f'(c)\Leftrightarrow \frac{\sqrt{y}-\sqrt{x}}{y-x}=\frac{1}{2\sqrt{c}}$

...

kittie21
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### Re: Going Crazy over this Mean Value Theorem Question!!!

...so.. the derivative of c, which is 1/2sqrt c should equal the slope of the tangent, sqrt y/ y. It makes sense in theory, but how do I make them equal eachother? Am I now supposed to solve for one of the variables? But there is two of them. I do not understand the next step. Thanks for your help so far though!

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
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### Re: Going Crazy over this Mean Value Theorem Question!!!

...so.. the derivative of c, which is 1/2sqrt c should equal the slope of the tangent, sqrt y/ y. It makes sense in theory, but how do I make them equal eachother? Am I now supposed to solve for one of the variables? But there is two of them. I do not understand the next step. Thanks for your help so far though!

$f'(t)=\frac{1}{2\sqrt{t}}$
$f'(c)=\frac{1}{2\sqrt{c}}$

$\frac{f(y)-f(x)}{y-x}=f'(c)\Leftrightarrow \frac{\sqrt{y}-\sqrt{x}}{y-x}=\frac{1}{2\sqrt{c}}$

...
$\frac{f(y)-f(x)}{y-x}=f'(c)\Leftrightarrow \frac{\sqrt{y}-\sqrt{x}}{y-x}=\frac{1}{2\sqrt{c}}\Leftrightarrow \sqrt{y}-\sqrt{x}=\frac{y-x}{2\sqrt{c}}$

you want to show that $\sqrt{y}-\sqrt{x}\leq\frac{y-x}{2\sqrt{x}}$

To get the above you need to answer for what value of $c$ is $\frac{y-x}{2\sqrt{c}}$ as large as possible...or
for what value of $c$ is $\sqrt{c}$ as small as possible.

remember...$0

kittie21
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### Re: Going Crazy over this Mean Value Theorem Question!!!

Soo... can I designate numbers to the value of x and y as long as they fit in with 0<x<y? How do I know which numbers to use to do that?

kittie21
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### Re: Going Crazy over this Mean Value Theorem Question!!!

okay so, the f(y)-f(x)/y-x has to be greater than 0, which means that b-a > 0 which means that the divisor must be positive so that the answer stays above zero, so b>a! I don't need to find an exact answer, do I?!

kittie21
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### Re: Going Crazy over this Mean Value Theorem Question!!!

I mean , so y>x...but i just realized that that is what the 0<x<y was saying anyways. nevermind. I am still confused!

Martingale
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### Re: Going Crazy over this Mean Value Theorem Question!!!

okay so, the f(y)-f(x)/y-x has to be greater than 0, which means that b-a > 0 which means that the divisor must be positive so that the answer stays above zero, so b>a! I don't need to find an exact answer, do I?!
I'm not sure what you are trying to accomplish with the above post.

I think at this point, if you don't see how to finish, you should look in your book to see examples of how the Mean Value theorem can be used. If I give one more hint I'll be doing the entire problem. (which doesn't really help you)