Question

how to find the equation of the line tangent to this

Answer 1

tangent to y= (2x+3)/(3-2) at point (1,5)

Answer 2

@FaiqRaees @jango_IN_DTOWN @zepdrix

Answer 3

by difference quotient formula or derivatives?

Answer 4

difference quotient

Answer 5

Okay start by finding the difference quotient of the function what do you get?

Answer 6

thats what i got

Answer 7

The formula you wrote above looks a little weird.
It should be,\(\large\rm \frac{f(x+h)-f(x)}{h}\) or \(\large\rm \frac{f(c+h)-f(c)}{h}\) if you're evaluating x at some point c.

But you shouldn't have x and c both in the expression.
Anyway, your set up looks correct :)

Now some annoying Algebra steps, ya?

Answer 8

yes i meant f(x)

Answer 9

So here is an option,\[\large\rm \frac{\dfrac{2(x+h)+3}{3(x+h)-2}-\dfrac{2x+3}{3x-2}}{h}\color{royalblue}{\left(\frac{[3(x+h)-2](3x-2)}{[3(x+h)-2](3x-2)}\right)}\]We can multiply through by the LCM of our denominator values.

Answer 10

Hmm your teacher is pretty cruel...
these problems are really pushing the boundaries of what you should be expected to do with the difference quotient. Teach made you expand a 4th degree polynomial.. and now this? Not cool...

Answer 11

ikr this is ap calc

Answer 12

Multiplying through by the LCM gives us,\[\large\rm \frac{[2(x+h)+3](3x-2)-(2x+3)[3(x+h)-2]}{h}\]Confused on that?

Answer 13

why not find derivative using quotient rule?

Answer 14

idkhow to thats y

Answer 15

\[(\frac{f}{g})' = \frac{f'g - fg'}{g^2}\]
f(x) = 2x +3 f'(x) = 2
g(x) = 3x-2 g'(x) = 3

Answer 16

okkk finally after all that i got 12xh+5h+12

Answer 17

thats what i got

Answer 18

omg im soooo sorrry oops

Answer 19

soo there was a little mistake that i catched

its supposed to be 12xh+5h/h

Answer 20

then idk what to do

Answer 21

yelp :(

Answer 22

Oh sorry I ran off to make a sammich :P
I didn't simplify it yet >.< ugh

Answer 23

well then you would factor out an "h" on top , then apply the limit h=0

However, that does not look correct

Answer 24

the question does not say how you have to find the derivative and I assume at this point in AP calc you have learned the following derivative properties (chain, product, and quotient)

Answer 25

She has not. Otherwise we would have done that...