(adsbygoogle = window.adsbygoogle || []).push({}); Question:Find an equation of the straight line that is tangent to the graph of f(x) = √x+1 and parallel tox-6y+4=0.

I've found f`(x).

When I write lim, assume "h->0" (as h approaches 0) is beneath it. I'd like to use LaTex to show you, but it seems too confusing to figure out.

f`(x)=lim f(x+h) - f(x) / h

f`(x)=lim √x+h+1 - √x+1/hx√x+h+1 + √x+1/√x+h+1 + √x+1

f`(x)=lim x+h+1-x-1/h(√x+h+1 + √x+1)

f`(x)=lim 1/h(√x+h+1 + √x+1)

f`(x)= 1/2(√x+1)

So that's f`(x). The problem is, I'm not sure where to go from here. I'm trying to find the equation of the line, and it has to be parallel to x-6y+4=0. Usually when I do a question like this, I'm given the point, but what do you do when you're given a line parallel to the tangent?

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# Derivatives - Finding Tangent

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