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 becomesy-sqrt3/4=-5sqrt3/24[x-3] and when simplified will give the answer above. Hope this was helpful.