Question

Differentiate y= square root of x - 1
Find the co=ordinates of the point on the curve at which the tangent is parallel to the line y=2-1.
Having trouble with this one, please help me!

Answer 1

So we differentiate it, and get \[y'=\frac{ 1 }{ 2\sqrt{x} }\] and what is the line? y=2x-1?

Answer 2

In any case, set the derivative = slope of the line and then solve for x

Answer 3

Yes, so i got slope of the line equal two. now I have \[\frac{ 1 }{ 2 \sqrt{x} }=2\] But I dont know how to go about solving that. Could you show me please?

Answer 4

At this stage you should know how to solve it, but in any case, start off by cross multiplying, what do you get?

Answer 5

\[\frac{ 2 }{ 2\sqrt{x} }\] then \[\frac{ 1 }{ \sqrt{x} }\]

Answer 6

\[\frac{ 1 }{ 2 \sqrt{x} }=2 \implies \frac{ 1 }{ 4 } = \sqrt{x}\] what's next?

Answer 7

Are you there?

Answer 8

Yes, is there a name for the solving of that so that I could brush up on some of the basic rules?

Answer 9

Hmm, I guess just google algebra, calculus is alllll algebra, so when we have square roots, we can square both sides to get rid of it, remember that square root(x) = x^(1/2) \[\sqrt{x} = \frac{ 1 }{ 4 } \implies (\sqrt{x})^2 = \left( \frac{ 1 }{ 4 } \right)^2 \implies x = \frac{ 1 }{ 16 }\]

Answer 10

\[\sqrt{x} = x^{1/2}\]

Answer 11

Oh ok. Cool thanks.

Answer 12

Yw

Answer 13

Here is khan academy it's great for learning things, also youtube has a lot of videos on algebra rules

https://www.khanacademy.org/math/algebra

Answer 14

Thanks :D