# Find the equation of the tangent line to the curve ..

• Dec 16th 2010, 11:47 AM
Find the equation of the tangent line to the curve ..
A) y = e^(-x) @ x = -1

dy/dx = (-1)(1)(e^(-x))
= -e^(-x)

f'(-1) = -0.368
f(-1) = 0.368

y =mx+b
0.368 = (-0.38)(-1) + b
0=b

y=-0.368x

B) y = 2 - e^(-x) @ x=1

dy/dx = 0 +(-1)(-e^(-x))
= e^(-x)

f'(1) = -0.368
f(1) = 1.632

y=mx+b
1.632 = (-0.38)(1)+b
2 = b

y= -0.368x + 2

Is this correct? I feel funny about it.
• Dec 16th 2010, 01:07 PM
dwsmith
1. \$\displaystyle m=-e \Rightarrow y-e=-e(x+1)\Rightarrow y=-ex\$

2. \$\displaystyle m=e^{-1}\Rightarrow y-(2-e^{-1})=e^{-1}(x-1)\Rightarrow y=e^{-1}x+2-2e^{-1}\$
• Dec 16th 2010, 02:04 PM
SammyS
Quote:

A) y = e^(-x) @ x = -1

dy/dx = (-1)(1)(e^(-x))
= -e^(-x)

f'(-1) = -0.368
f(-1) = 0.368 , no, f(-1)=e, and f'(-1)=-e

y =mx+b
0.368 = (-0.38)(-1) + b
0=b

y=-0.368x

B) y = 2 - e^(-x) @ x=1

dy/dx = 0 +(-1)(-e^(-x))
= e^(-x)

f'(1) = -0.368 : This should be e^(-x) for x=1, so it's positive.
f(1) = 1.632 : This is correct.

y=mx+b
1.632 = (-0.38)(1)+b
2 = b

y= -0.368x + 2

Is this correct? I feel funny about it.

For (A):

\$\displaystyle \displaystyle e^{-x}\$ at x=-1 is: \$\displaystyle \displaystyle e^{-(-1)}=e^1=e\approx 2.718281828\$

Otherwise, what you did was right. (You should be able to check these with a graphing calculator.)

For (B):

\$\displaystyle \displaystyle e^{-(1)}=e^{-1}\approx 0.36788\$