let f(x) be defined as f(x)=1+2x-x^3. Find the equation of the line tangent to f(x) in slope-interecpt form at the point where x=1
f'(x) = 2-3x m = 2-3(1) = -1 y = 1 + 2(1) - (1)^3 = 2 (1,2) Tangent: m = -1 y - y^1 = m (x-x^1) y - 2 = -1(x-1) y - 2 = -x + 1 T: x - y -3 = 0 Like that ? xD
i think your answer is right, but the derivative is \[f'(x)=2-3x^2\]
Woops , but it's right anyway since x is 1
LOL
Shut up xD People makes mistakes that was a typo :P
i know it was a typo. i make them all the time.
can u tell what step goes first, and thereafter
Hm , Saifoo can explain :P
lol, im learning that.. i learned some of it yesterday
Learning it!? Yeah right, I thought you passed HS?
HS here is a bit different. :P
we pass HS in 10th. then study 2 years college after it.
wow~ in yr 10!? :O:O So yr 11 and 12 is college there ?
oh yes this is very helpful <_< lol nevermind. i found my notes. this is too easy lol
LOL, sorry need me to elaborate ?
if u dont mind
yessss!!
so the final answer should be -1x+3 right?
Okay since Saifoo can't help :P First you have to find the derivative then it's f'(x) = 2 - 3x^2 Then you have to find the gradient using x = 1 , f'(x) = 2 - 3(1)^2 m = -1 Oh yeah then find the y by substituting x into the thing y = 1 + 2(1) -(1)^3 = 2 Tangent: (x1, y1) = (1,2) y - y1 = m(x-x1) y - 2 = -1(x-1) y - 2 = -x + 1 T; x - y -3 = 0
y = x - 3 , if its not in general form
hold on i got -1x+3
y=m(x-x1)+y -1(x-1)+2
what? I don't get yur way
well my way is basically ur way except i add y to the other side. and when siimplified it should be -x+3, your doing something wrong
T; x - y -3 = 0 where does this come from?
I doubled check, nothing is wrong =/
Oh, we did it in different forms, your answer is right, my answer is right as well
ok thx
yw (:
I did do a stupid mistake sorry =/ T: x+y-3 = 0
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