Question

What is the tangent line of y = sqrt(x-1)? Please show steps thanks

Answer 1

you can only get the equation of the gradient of the tangent from differentiating. You will need a point or ordered pair to find the equation of the tangent.

rewrite the equation in index form

\[y = (x + 1)^{\frac{1}{2}}\]

the differentiating using the chain rule

\[\frac{dy}{dx} = \frac{1}{2} ( x + 1)^{-\frac{1}{2}}\]

or

\[\frac{dy}{dx} = \frac{1}{2\sqrt{x + 1}}\]

but this is only the equation of the gradient of the tangent.... you still need an ordered pair or x value for a specific tangent

Answer 2

ok for the ordered pair (2,1)

Answer 3

ok to find the gradient of the tangent at x = 2 just substitute x =2

\[m = \frac{1}{2\sqrt{2 + 1}}... m = \frac{1}{2\sqrt{3}}\]

then using the point slope formula

\[y - 1 = \frac{1}{2\sqrt{3}}( x - 2)\]

will be the equation of the tangent at x = 2

Answer 4

oops you'll need to simplify things a bit

Answer 5

So I can just do the power rule to find the f^1(x) and that will give me the slope?

Answer 6

it gives the equation of the slope... you still need an x value to get the slope at a point...

Answer 7

ok and once you have the slope how do you put it into the linear form? can you do y = mx+b?

Answer 8

yes you can... once you know m.... put it into slope intercept form.. substitute the ordered pair to find the intercept.

Answer 9

so is the slope at (2,1) 2?

Answer 10

no its \[m = \frac{1}{2\sqrt{3}}\]

this comes from substituting x = 2 into the 1st derivative

Answer 11

why is it 1/2(x+1)^(-1/2) instead of 1/2(x-1)^(-1/2) if the original function was
f(x) = (x-1)^(-1/2)?

Answer 12

the original function you posted was

\[y = \sqrt{x + 1}... or... y = (x+1)^{\frac{1}{2}}\]

sorry but I have to go...

good luck with it

Answer 13

no I posted y = sqrt(x-1) but okay