using f(x)=1/sqrtx, find a tangent line that is parallel to x+2y+6=0

Question

I can get most of the way through this problem.  I know the tangent line is y=-1/2x+3/2.  I know it intersects the function at (1, 1).  I get stuck when trying to prove it.  using the formula:

 

limit  of (f(c+deltax)-f(c))/deltax as deltax approaches 0

 

I put my funciton into the formula and remove the fraction from the top.  Then, I multiply by the conjugate.  Then, i Simplify and I am left with:

 

-1/(sqrtx)(sqrt(x+deltax))(sqrtx+sqrt(x+deltax))

 

That is where I am stuck.  So, my questions is:  How do I get to (y=-1/2x+3/2) from here?  As always please be as specific as possible.  Thank you so much for your help.

Answer 1

First, let us find the slope of x+2y+6=0:

 

y = -(1/2)x - 3

 

Slope is m = -1/2

 

Next, set dy/dx of the given function equal to m.  That will be the slope of a line parallel to x+2y+6=0:

 

y = 1 / sqrt(x)

 

dy/dx = -(1/2) / x^(3/2) = m = -1/2

 

x^(3/2) = 1, so x = 1, and f(1) = 1 / sqrt(1) = 1 (note x>0, so we can neglect the negative square root)

 

So far, the tangent point is (1,1) and the slope of the tangent line is m = -1/2.  Now put this in form y=mx+b and find b:

 

1 = -(1/2)(1) + b, so b = 3/2

 

Final equation of tangent line at (1,1):

 

y = -(1/2)x + 3/2

 

I think you had some difficulty evaluating the derivative of 1/sqrt(x).  Just use the form:

 

y = x^k with  k = -1/2

 

dy/dx = kx^(k-1) = -(1/2)x^(-1/2 - 1) = -(1/2) / x^(3/2)