Question

evaluate the tangent line to the curve

Answer 1

first things first.

This is what I understand so far

See I understand now that the limit is the instantaneous slope

You take say one point set.

x, f(x)

then you go some small distance h

x+h, f(x+h)

so you find the slope of these two points.

\[\frac{ f(x+h)-f(x) }{ x+h-x } = \frac{ f(x+h)-f(x) }{ h } \]

so I guess the value of this as the function gets closer to zero or the limit of this would be equal to the instantaneous slope, or derivative.

\[\lim_{h \rightarrow 0} \frac{ f(x+h)-f(x) }{ h }\]

Answer 2

now my question is say we have a function

\[y = \frac{ x+1 }{ x-3 }\]

how would we find say the tangent line to the curve above at the point (5,3) ?

I know they are asking for the derivative of this function but how would we do this using

\[\lim_{x \rightarrow 0} \frac{ f(x+h)-f(x) }{ h }\]

Answer 3

It's actually h approaches to 0 (in the 2nd response), but it's still pretty straightforward.
\[f(x) = \frac{x+1}{x-3}\]therefore\[f(x+h)=\frac{(x+h)+1}{(x+h)-3}\]so far so good?

Answer 4

okay, so you just inserted (x+h) wherever you saw x

Answer 5

Yea, x is just a variable, so i could put whatever i want instead of x. In this case x+h. Now substitute the rest, meaning finish off\[\lim_{h->0}\frac{f(x+h)-f(x)}{h}\]

Answer 6

Where\[f(x+h) = f(5+h) = \frac{(5+h)+1}{(5+h)-3}\]and\[f(x)=f(5)=3\]

Answer 7

Now you're just left with limit as h approaches 0 of an expression that's only in terms of h, so just take that limit and you're set. Does this make sense?

Answer 8

Yes so far so good let me show you my work so far

Answer 9

This is what I did

\[\frac{ \frac{ 5+h+1 }{ 5+h-3 } -\frac{ 5+1 }{ 5-3} }{ h} = \frac{ \frac{ 5+h+1 }{ 5+h-3 } -3}{ h}\]

\[\frac{ \frac{ 6+h }{ 2+h }*(2+h) -3(2+h)}{ h(2+h)} = \frac{ (6+h)-3(2+h) }{ h(2+h) }\]

Answer 10

Looks good, continue simplifying.

Answer 11

\[\frac{ (6+h)-(6+3h) }{ h(2+h) } = \frac{ -2h }{ h(2+h) } = \frac{ -2 }{ 2+h }\]

\[|_{h = 0} \frac{ -2 }{ 2+h } = \frac{ -2 }{ 2 } = 1 \]

Answer 12

-1 **

Answer 13

Sorry crashed, but yea looks right to me (-1)

Answer 14

Thanks for your guidance! I have another one i'm not too sure about

Answer 15

Sure

Answer 16

i'll tag you in another question, it shouldn't be that long

Answer 17

ok sure