tangent line curve

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• Apr 17th 2012, 01:00 AM
tankertert
tangent line curve
Hi guys.

equation of tangent line to the curve $y = 3x^2 + e^{1-x}$ when x=1

i tried and got 0

can you show me how to solve with working please.
thank you!
• Apr 17th 2012, 01:29 AM
biffboy
Re: tangent line curve
dy/dx=6x-1 So when x=1 gradient of tangent=5 Also when x=1 y=2+e

So you want a straight line with gradient 5 through the point (1,2+e)
• Apr 17th 2012, 01:42 AM
a tutor
Re: tangent line curve
Quote:

Originally Posted by tankertert

equation of tangent line to the curve $y = 3x^2 + e^1-x$ when x=1

I suspect OP meant $y = 3x^2 + e^{1-x}$.

If so OP remember to write it like this e^{1-x}.
• Apr 17th 2012, 03:02 AM
tankertert
Re: tangent line curve
Quote:

Originally Posted by a tutor
I suspect OP meant $y = 3x^2 + e^{1-x}$.

If so OP remember to write it like this e^{1-x}.

sorry yes that what i mena
• Apr 17th 2012, 03:32 AM
biffboy
Re: tangent line curve
So now dy/dx=6x-e^1-x When x=1 y=4 and dy/dx=5 Require straight line through (1,4) with gradient 5
• Apr 17th 2012, 09:47 PM
tankertert
Re: tangent line curve
Quote:

Originally Posted by biffboy
So now dy/dx=6x-e^1-x When x=1 y=4 and dy/dx=5 Require straight line through (1,4) with gradient 5

but how did you get that