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Mathematics 18 Online
OpenStudy (anonymous):

can some one pls explain this to me: The limit as h goes to 0: ((sqrt(3+h)-sqrt(3)))/h represents the derivative of a function f at a number c. Determine f and c. Thank you!

OpenStudy (amistre64):

sqrt(x) derives to 1/2sqrt(x)

OpenStudy (amistre64):

\[\frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{\sqrt{x+h}}{h}-\frac{\sqrt{x}}{h}\] cant recall it really

OpenStudy (amistre64):

maybe it the \[\frac{f(b)-f(a)}{b-a}\]version

OpenStudy (amistre64):

ueah; i think its has to do with the conjugate

OpenStudy (amistre64):

\[\frac{\sqrt{x+h}-\sqrt{x}}{h}*\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\]

OpenStudy (amistre64):

tryig to type and math at the same time... messed it up lol

OpenStudy (anonymous):

haha - no worries< Im going through it all

OpenStudy (anonymous):

what does the q mean when it asks for f and c? that;s my real issue

OpenStudy (amistre64):

that 2sqrt(garbage) up top is not spose to even be there; when you multiply conjugates there is no middle term :)

OpenStudy (amistre64):

q?

OpenStudy (anonymous):

question \

OpenStudy (amistre64):

oh; its asking for the derivative of the function and the value used which is 3 n this case

OpenStudy (anonymous):

I calc'ed it hoping that;s what it meant :-)

OpenStudy (anonymous):

what do you mean by value?

OpenStudy (amistre64):

\[\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\] \[\lim_{h->0}\ \frac{1}{\sqrt{x+h}+\sqrt{x}}\ =\frac{1}{\sqrt{x}+\sqrt{x}}\]

OpenStudy (amistre64):

you see how they begin with 3+h instead of x+h? c = 3

OpenStudy (amistre64):

f(c) = \(\frac{1}{2\sqrt{c}}\) when c = 3

OpenStudy (anonymous):

yes, i see that

OpenStudy (anonymous):

so c is the x value?

OpenStudy (amistre64):

yes; c usually denotes the term of the limit as in \(\lim_{x->c}\)

OpenStudy (anonymous):

oh, I see!

OpenStudy (anonymous):

to determine the f or derivative would I be required to plug in the 0 at the end? since it is h goes to 0?

OpenStudy (anonymous):

c = ssqrt3

OpenStudy (amistre64):

\[\lim_{x->3}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{3}}\]

OpenStudy (anonymous):

here's the answer

OpenStudy (anonymous):

f(x)=sqrtx at the point c = 3

OpenStudy (amistre64):

well, yeah; sqrt(x) derives to; 1/2sqrt(x)

OpenStudy (amistre64):

i try to do those mathlab type websites where they give you practice problems ...... i loathe them. they never seem to be able to say exactly what it is they want you to do lol

OpenStudy (anonymous):

that was my issue with this question....

OpenStudy (anonymous):

Thanks for your help!

OpenStudy (amistre64):

:) just swear at the computer monitor and click next :)

OpenStudy (anonymous):

so to calc f I am saying as x goes to 3

OpenStudy (anonymous):

hahahaha! I wish I could. I am changing careers and they wont take my highschool calc classes ( over a decade ago) so im back in calc and have to do well

OpenStudy (amistre64):

the question asked, in hindsight, what function are they deriving and what did they use for 'c' f(x) = sqrt(x) and c = 3

OpenStudy (anonymous):

thanks!

OpenStudy (anonymous):

that was a really backwards way to ask such an easy question?

OpenStudy (anonymous):

thanks a ton amistre! you are awesome! off to sleep, have to wake up early to trade oil :-)

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