How do I find an equation for the tangent line to the graph of #f(x)=(sqrt(x))/(4x-8)# at the point #(3, f(3)#)?

I don't understand what I'm supposed to do with #f(3)# for #y#. Am I supposed to find the derivative of the graph for the slope at the points first then somehow use x and y to find my #y=mx+b#?

1 Answer
May 12, 2018

Equation of tangent line , #y=[[-5sqrt3][ x]]/24+[21sqrt3]/24#

Explanation:

To find the equation of the tangent line we must first find the gradient at the given point, i.e, #x=3# To do this we have to differentiate the expression #f[x]=sqrtx/[4x-8]# using the the quotient rule, #d[u/v]= [vdu-udv]/[v^2]# where u and v are functions of #x#. [# v=4x-8 and u=sqrtx# in this case.]

So , #d/dx[sqrtx/[4x-8]]#=#[4x-8]d/dxsqrtx-sqrtxd/dx[4x-8]#

#f'[x]#=#[[4x-8]/[2sqrtx]-4sqrtx]/[[4x-8]^2#, if you substitute #x=3# into this expression this will give a gradient of #-5sqrt3/24# and #f[3]#= #sqrt3/4#.
And from the equation of a straight line ,#[y-y1]=m[x-x1]# where #m# is the gradient,

So the equation of the tangent line becomes#y-sqrt3/4=-5sqrt3/24[x-3]# and when simplified will give the answer above. Hope this was helpful.