Question

Let f(x) = sqrt {1-x^2}, A + (-1,f(-1)) and B = (1,f(1)). Find a tangent to f in the interval (-1,1) that is parallel to the secant AB.

Answer 1

* A= (-1, f(-1))

Answer 2

OK, so A is at (-1,0) and B is at (1,0).
The secant line through A and B is horizontal and on the x-axis.

Answer 3

The slope of secant AB is zero, so your tangent line also must have a tangent of zero to be parallel.
f(x) is the top half of a semi-circle centered at the origin, so the tangent will occur at its top, at x=0.

Answer 4

Let me know if you need any more help. I think you can get it from here.

Answer 5

[correction: "so your tangent line also must have a tangent of zero . . " should read "... have a *slope* of zero . . "

Answer 6

So I set the derivative of the function = 0?