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secant line, tangent line

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f(x) = sqrt(x-1), 1<=x<=3
Let A = (a,f(a)), and B = (b,f(b)).
Write an equation for
1. the secant line AB
2. a tangent line to f in the interval (a,b) that is parallel to AB

Thank you for all the help.

For the line, you have two points,
y=mx+b put the two points in, and solve for the two unknowns, m, and b. Two equations, two unknowns.

For 2) Take the derivative, set it equal to the slope m in the above solution, solve for x, then, knowing f(x), solve for y (f(x)).

I will be happy to critique your work.

Thanks. I did this for f(x) = x + 1/x. 0.5<=x<=2

I got the points as A(.5,2.5) and B(2,2.5)
Plugged it in y=mx+b, and got
m=0 & b=2.5

So the eq. is y = 2.5

For 2, I took derivative and solved for x. I got x = +1, -1

So y = 1+1/1 = 2

Do we ignore -1?

Ignore it. The original function was
f(x) = sqrt(x-1), 1<=x<=3 , so x=-1 is outside the domain.

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