could some check solution please

**decoy808** · February 17th 2010, 03:18 PM

Given the function

$y= \frac{x+1}{x}$

a. Find the equation of the tangent to the curve at the point (2,(3/2))

b. Find the value of x where the tangent to the curve is
perpendicular to the tangent found in part (a)

$<br /> \frac{x+1}{x}= \frac{1}{x} - \frac{x+1}{x^2}$

$<br /> \frac{1}{2} - \frac{2+1}{2^2}$

m= 1/4

y-y1=m(x-x1)
y- (3/2) = 1/4(x-2)
y= 1/4x +1

part B

M1,M2 = -1
1/4 M2 = -1
m2 = 4

y- (3/2) = 4(x-2)
y=4x-6.5

if some could confirm iif ihave done things correct mucho gratis

**pickslides** · February 17th 2010, 03:30 PM

G'day, I think your on the right track,

I would say $y= \frac{x+1}{x} = 1+\frac{1}{x}$ then $y' = \frac{-1}{x^2}$

Then find $y'(2) = \frac{-1}{2^2}=\frac{-1}{4}$ to make $m$ in the tangent equation.

**Chris11** · February 17th 2010, 10:46 PM

yeah, it looks good

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