Find using the defintion f '(c) lim = f(x) - f(c)

x -> c _______

x - c

the gradient function f ' (x) for:

a) f(x) = 1/x

b)f(x) =√ x

[My tutor told me to rationalize before doing the question for part b but how do i rationalize this]

- Jul 19th 2008, 07:58 AMsweetGFind using the definition........the gradient function f '(x)
**Find using the defintion f '(c) lim = f(x) - f(c)**

x -> c _______

x - c

the gradient function f ' (x) for:

a) f(x) = 1/x

b)f(x) =**√ x**

[My tutor told me to rationalize before doing the question for part b but how do i rationalize this]

- Jul 19th 2008, 08:15 AMflyingsquirrel
Hi

To rationalize the numerator of $\displaystyle \frac{\sqrt{x}-\sqrt{c}}{x-c}$, multiply and divide the fraction by the conjugate of $\displaystyle \sqrt{x}-\sqrt{c}$ which is $\displaystyle \sqrt{x}+\sqrt{c}$ :

$\displaystyle \frac{\sqrt{x}-\sqrt{c}}{x-c}=\frac{(\sqrt{x}-\sqrt{c})(\sqrt{x}+\sqrt{c})}{(x-c)(\sqrt{x}+\sqrt{c})}=\ldots$

(notice that the numerator looks like $\displaystyle (a-b)(a+b)$ which equals $\displaystyle a^2-b^2$)

Edit : for the first question : $\displaystyle \frac{f(x)-f(c)}{x-c}=\frac{\frac{1}{x}-\frac{1}{c}}{x-c}=\frac{\frac{c-x}{xc}}{x-c}=\frac{c-x}{xc(x-c)}=-\frac{1}{xc}$ hence $\displaystyle f'(c) = ?$ - Jul 20th 2008, 05:51 AMsweetG
Sorry I didn't realize that the definition for f '(c) was all over the place....sorry about that.

For Part b is the answer:

**(x-c)/(x****√x + x√c - c√x - c√c**and for Part (a) how am I supposed to find f '(x)

Sorry this question is really confusing me

coz there are so many x's and c's.......

- Jul 20th 2008, 06:40 AMflyingsquirrel
That's it but if you try computing $\displaystyle f'(c)=\lim_{x\to c}\frac{x-c}{x\sqrt{x}+x\sqrt{c}-c\sqrt{x}-c\sqrt{c}}$ you'll see that this limit is an indeterminate form. (both the numerator and the denominator tend to 0) If you hadn't expanded the numerator, it would have given you

$\displaystyle \frac{\sqrt{x}-\sqrt{c}}{x-c}=\frac{(\sqrt{x}-\sqrt{c})(\sqrt{x}+\sqrt{c})}{(x-c)(\sqrt{x}+\sqrt{c})}=\frac{x-c}{(x-c)\left(\sqrt{x}+\sqrt{c}\right)}=\frac{1}{\sqrt{x }+\sqrt{c}}$

And $\displaystyle f'(c)=\lim_{x\to c}\frac{1}{\sqrt{x}+\sqrt{c}}$ isn't an indeterminate form any longer. Can you find the value of this limit ?

Quote:

and for Part (a) how am I supposed to find f '(x)

Sorry this question is really confusing me

coz there are so many x's and c's.......

- Jul 20th 2008, 08:01 AMsweetG
Does f '(c) = 1/2√c

which therefore makes f ' (x) = 1/2√x

Is this right or have I made a mistake ? - Jul 20th 2008, 10:52 AMflyingsquirrel