Question

Can someone check my work?
What is the tangent line to the curve at the given point.
y=sqrt(x) (1,1)

My answer is...
f(x)-f(a)/x-a
sqrt(x)-1/1-1=sqrt(x)-1/0


is this right? If not, can you correct me?

Answer 1

I don't think that's quite it... looking at it now though

Answer 2

differentiate it \[\frac{dy}{dx}=\frac{1}{2\sqrt{x}}\]
then sub in \(1\)

Answer 3

yes, thanks! my equation typing energy was running low :)

Answer 4

1/2? is it y=1/2x+3/2?

Answer 5

y=1/2x-3/2 correction

Answer 6

the derivative of the curve gives the slope of the curve

Answer 7

I know that.. which means the slope is 1/2

Answer 8

so you want the slope of the curve f(x) at point (1,1), so you put x = 1 into the dy/dx expression and get 1/2

Answer 9

that's the slope. Then use that same point to build the y = mx + b line equation for the tangent

Answer 10

y=1/2x+1/2?

Answer 11

yeah I think that's better... good fix.

Answer 12

thanks.

Answer 13

glad to help :)