Question

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`?

Answer 1

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.