An interesting question.
Find the equation of the tangent for the curve at the point without using calculus.
you mean without using differentiation, right? well, if so, suppose the line is since (1,1) belongs to the line, we have now the tangent line should intersect the curve
at one point only, i.e. the equation must have only one solution. so the discriminant has to be 0, which gives us thus the line is
1. at . We know the tangent line is
Ok so taking a cue from NonCommAlg and assuming that you mean no derivatives this is simple. It can be easily show using non caluculus terms that the function approaches one as x approaches zero. So now consider this let be a differentiable function and its tangent line at a point . Consider that a tangent line approximates a curve in a neighborhood of , and that . So it is obvious that must get closer and closer to one as x gets closer and closer to c. So we must have that must get closer and closer to one as x gets closer and closer to zero.But because as we noted we must have that so the tangent is .
I implictly used calculus I suppose.
I think for the second one I can do better
For the function we know that the tangent line, , at must be of the form . Now consider the tangent line of the function at . It is clear geometrically that these functions have perpendicular slopes, so the tangent line must be of the form . But now consider that in this case so
Im not too sure about that one...can someone verify?