need help
\[f(x) = \frac{(2x+3)}{(3x-2)}\] we need to find its derivative
we also need to determine the value of f(1)
yes, but how would take the derivative with that function?
by using what is so adequately refered to as: the quotient rule
Do you know the derivative rules yet?
we could also get by with bringing the bottom to the top and placing a negative exponent on it and derive it with the product rule
Well we haven't cover the quotient rule yet, but if you could use it to show me I would appreciate because I believe my professor was going to cover, but ran out of time
\[\frac{top}{bottom}\] \[Dx\ (\frac{t}{b}) = \frac{bt'-b't}{b^2}\]
does that make sense as written?
yes
good, find the derivates of the top and the bottom, then plug them into the right spots in the formula
keep attantion to your signs; youve got some mismatched
its good form except for you should have (3x-2), not (3x+2).
unless of course i mistyped the original :)
when you have it all squared away, find f'(1) by using x=1 in the derivative to find the value of the slope of the tangent line at f(1)
I get f(1)=-13
what do I do next?
i got -14 .... you sure?
lol ... its -13
that is the slope of the tangent line; use it in the slope point form of an equation
y-f(1) = f'(1)(x-1)
I got y=-13x+18
did you get the same?
y-f(1) = f'(1)(x-1) y-5 = -13x +13 .. yes
yes
whats next?
thats it, you found the equation of the tangent line at 1,f(1)
just match it to one of the formats they present
id go with C myself
But c has a positive 13 in front of the x and our answer has -13 front of the x
y = -13x + 18 +13x +13x ----------------- 13x+y = 18
you have to move stuff around to match what they present
Thank you!
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