1. ## tangent problem

An interesting question.

Find the equation of the tangent for the curve $y = x^2$ at the point $(1,1)$ without using calculus.

2. Originally Posted by danny arrigo
An interesting question.

Find the equation of the tangent for the curve $y = x^2$ at the point $(1,1)$ without using calculus.
you mean without using differentiation, right? well, if so, suppose the line is $y=mx + b.$ since (1,1) belongs to the line, we have $b=1-m.$ now the tangent line should intersect the curve

at one point only, i.e. the equation $x^2 - mx + m-1=0$ must have only one solution. so the discriminant has to be 0, which gives us $m=2.$ thus the line is $y=2x-1.$

3. Okay, so how about $y = x^n$ say for some positive integer $n$?

4. Originally Posted by danny arrigo
Okay, so how about $y = x^n$ say for some positive integer $n$?
since x = 1 must be the root of the equation $x^n-mx+m-1=0$ of multiplicity at least 2, we'll have: $x^n - mx + m - 1=(x-1)^2p(x),$ for some polynomial $p(x).$ thus:

$(x-1)(x^{n-1}+ \cdots + x + 1 - m)=(x-1)^2p(x),$ and hence: $x^{n-1} + \cdots + x + 1 - m = (x-1)p(x).$ not put $x=1$ to get $m=n,$ and therefore $b=1-n.$ so the equation

of the tangent line is $y=nx + 1-n.$

5. $y = \sin x$ or $y = e^x$ at say $(0,0)$ or $(0,1)$, resp.?

6. Originally Posted by danny arrigo
$y = \sin x$ or $y = e^x$ at say $(0,0)$ or $(0,1)$, resp.?
Why do you think they invented Calculus!

7. Originally Posted by danny arrigo
$y = \sin x$ or $y = e^x$ at say $(0,0)$ or $(0,1)$, resp.?
1. $\sin(x)$ at $(0,0)$. We know the tangent line is $\hat{y}-0=\cos(0)(x-0)\implies \hat{y}=x$

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 $\frac{\sin(x)}{x}$ approaches one as x approaches zero. So now consider this let $f$ be a differentiable function and $\hat{y}$ its tangent line at a point $x=c$. Consider that a tangent line approximates a curve in a neighborhood of $c$, and that $\hat{y}(c)=f(c)$. So it is obvious that $\frac{f(x)}{\hat{y}}$ must get closer and closer to one as x gets closer and closer to c. So we must have that $\frac{\sin(x)}{mx}$ must get closer and closer to one as x gets closer and closer to zero.But because as we noted $\frac{\sin(x)}{x}\to 1$ we must have that $\frac{\sin(x)}{mx}\to 1 \implies m=1$ so the tangent is $y=x$.

I implictly used calculus I suppose.

I think for the second one I can do better

For the function $f(x)=e^x$we know that the tangent line, $\hat{y}$, at $(0,1)$ must be of the form $\hat{y}-1=mx\Rightarrow \hat{y}=mx+1$. Now consider the tangent line $\hat{y}_2$ of the function $f_1(x)=e^{-x}$ at $(0,1)$. It is clear geometrically that these functions have perpendicular slopes, so the tangent line must be of the form $\hat{y}_2=\frac{-x}{m}+1$. But now consider that $f_1(-x)=f(x)$ in this case so $\hat{y}(x)=mx+1=\frac{x}{m}+1=\hat{y}_2(-x)\implies m=1$

Im not too sure about that one...can someone verify?