Question

why is the slope of 1/x-1 at x=2

1?

Answer 1

It is a horizontal translation of 1/x, shifted to the right one unit. 1/x has a slope of 1 at x = 1 so this function has that slope at x = 2.

Answer 2

Or you could try the definition of a derivative...

Answer 3

\[\frac{ 1 }{ x-1 } \] at x = 2, so it's translated down 1 unit.

Answer 4

i'm trying to find the slope of the curve

Answer 5

do you know how to find the derivative?
if not, use
\[ \frac{f(x+h)-f(x)}{h} \]
and let h->0 to find the slope. then plug in x=2 to get the value of the slope

Answer 6

\[ \frac{f(\frac{ 1 }{ 2-1 }) +h - f(2) }{ h }\]

\[\frac{ 1+h-1}{ h} = 1\]

Answer 7

i do know how to do the derivative... i was really unsure lol

Answer 8

in this case
\[ f(x)= \frac{1}{x-1} \]
and
\[ f(x+h)= \frac{1}{(x+h)-1}\]
can you finish?

Answer 9

\[\lim_{h \rightarrow 0}\frac{ 1 }{ h}\left( \frac{ 1 }{ x+h-1 } -\frac{ 1 }{ x-1 } \right)\]

Answer 10

well i got one either way. is my way correct or wrong?

Answer 11

But the answer is not 1.

to finish
\[ \lim_{h \rightarrow 0}\frac{ 1 }{ h}\left( \frac{ x-1 }{ (x-1)(x+h-1) } -\frac{ x+h-1 }{ (x-1)(x+h-1) } \right)\]

\[ \lim_{h \rightarrow 0}\frac{ 1 }{ h}\left( \frac{ -h }{ (x-1)(x+h-1) } \right)\]
\[ \lim_{h \rightarrow 0}\left( \frac{ -1 }{ (x-1)(x+h-1) } \right)\]
now let h go to zero to get
\[ \frac{ -1 }{ (x-1)(x-1) } \]
or
\[ f'(x)= \frac{-1}{(x-1)^2} \]

plug in x=2 to find the slope at x=2
you get -1

Answer 12

once you learn derivatives
\[ f(x)= (x-1)^{-1} \]
and
\[ f'(x)= -(x-1)^{-2} \]
follows immediately from the rule on differentiating powers.