# pre-calculus/functions help

• Oct 1st 2009, 03:51 AM
str33tl0rd
pre-calculus/functions help
hello all,

question:

$f(x)=1/x$ evaluate $f(x)-f(a)/x-a$

thank you
• Oct 1st 2009, 06:05 AM
ialbrekht
$f(x)-f(a)/x-a = 1/x-(1/a)*(1/x)-a = f(x)(1-1/a)-a$

or you can write it as

$f(x)-f(a)/x-a = 1/x-(1/a)*(1/x)-a = (a-1)/(a*x)-a$
• Oct 1st 2009, 06:06 AM
VonNemo19
Quote:

Originally Posted by str33tl0rd
hello all,

question:

$f(x)=1/x$ evaluate $f(x)-f(a)/x-a$

thank you

There is nowhere to start. It looks like you've got the slope of a secant line through a hyperbola (assuming that you meant to include parentheses).

Do you have any more instruction?
• Oct 1st 2009, 05:08 PM
str33tl0rd
thank you for the help guy, but it seems like that wasn't the answer =|

i think i have miswritten the question, i'll retry:

if $f(x)=1/x$
evaluate
$[f(x)-f(a)]/(x-a)$

and the answer is supposed to be: $-1/ax$

i need a working out, thank you very much all you guys and girls (Wink)
• Oct 1st 2009, 05:30 PM
skeeter
Quote:

Originally Posted by str33tl0rd
thank you for the help guy, but it seems like that wasn't the answer =|

i think i have miswritten the question, i'll retry:

if $f(x)=1/x$
evaluate
$[f(x)-f(a)]/(x-a)$

and the answer is supposed to be: $-1/ax$

$\frac{\frac{1}{x} - \frac{1}{a}}{x-a}$

get a common denominator for the two terms in the numerator and combine into a single fraction ...

$\frac{\frac{a-x}{ax}}{x-a}$

now finish it
• Oct 1st 2009, 05:51 PM
str33tl0rd
thank you, thats the step i got up to:

$

\frac{\frac{a-x}{ax}}{x-a}
$

and i continued as follows, correct me if i am wrong:
$\frac{a-x}{ax*(x-a)}$
expanding:
$\frac{a-x}{ax(sqr)-a(sqr)x}$

and here is where i get stuck, don't know how to continue or simply that, thanks again
• Oct 1st 2009, 05:54 PM
skeeter
Quote:

Originally Posted by str33tl0rd
thank you, thats the step i got up to:

$

\frac{\frac{a-x}{ax}}{x-a}
$

and i continued as follows, correct me if i am wrong:
$\frac{a-x}{ax*(x-a)}$

do not $\to$ expand

now, what does $\textcolor{red}{\frac{a-x}{x-a}}$ = ???

...
• Oct 1st 2009, 05:58 PM
str33tl0rd
thank you very much...(Rock)