tangent line curve

• Apr 17th 2012, 12:00 AM
tankertert
tangent line curve
Hi guys.

equation of tangent line to the curve \$\displaystyle 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, 12:29 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, 12:42 AM
a tutor
Re: tangent line curve
Quote:

Originally Posted by tankertert

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

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

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

Originally Posted by a tutor
I suspect OP meant \$\displaystyle 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, 02:32 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, 08:47 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