Find the equation of the tangent to the curve y= e^-x which

[kimmy_koo51]

Find the equation of the tangent to the curve y= e^-x which is parallel to 2y+x=0

y = e^-x
y' = -e^-x

2y= -x
y= -1/2x

Since the lines are parallel the slopes are the same

-e^-x = -1/2
x log e = log -1/2
x= (log -1/2)/log e
x= 0.69

y= e^-x
y= e^ -(0.69)
y= 0.5

y - 0.5 = 1/2 (x-0.69)
2y - 1 = x - 0.69
x - 2y + 1 = 0

Is this correct...it just doesn't feel right for some reason.

 

[galactus]

\(\displaystyle {-}e^{-x}=\frac{-1}{2}\)

\(\displaystyle x=ln(2)\)


\(\displaystyle \L\\y=mx+b\)

You know the slope is -1/2


\(\displaystyle \L\\e^{-ln(2)}+\frac{1}{2}ln(2)+b\)

\(\displaystyle \L\\b=\frac{1}{2}(1+ln(2))\)

Therefore, the line equation is:

\(\displaystyle \L\\y=\frac{-1}{2}x+\frac{1}{2}(1+ln(2))\)

 

[pka]

\(\displaystyle \L\begin{array}{rcl}
- e^x & = & - \frac{1}{2} \\
e^x & = & \frac{1}{2} \\
x & = & \ln \left( {\frac{1}{2}} \right) \\
\end{array}.\)

 

[kimmy_koo51]

Thanks a lot you guys...that makes this a whole lot easier.