On the curve $y =\frac{ 2x - 1}
{x - 1}$ , $x > 1$ drag tangent such that area of right triangle bounded by this tangent and the axes is minimal .
What is the minimum value of this area ?
My work:
Let $y=kx+l$ be that tangent line in the first quadrant.
We know that line goes through points $(0,l)$ and $(-\frac{l}{k},0).$ So area is $A=\frac{l*-\frac{l}{k}}{2}=-\frac{-l^2}{2k}$
But area depends of two variables and we want to depends on just one because we will than solving equation: $f'(l)$ or $f'(k)$.
Let rewrite our function: $f(x)=\frac{1}{x-1}+2$ Than we can easily see that $f'(x)=-\frac{1}{(x-1)^2}$. That is our k and from equation $y=kx+l$ we will get $l$ and than we can solve eqatuion $A'(x)=0$ to check in which point area is max. And after we know point we can easily know area of triangle.
Please solve by just using differentiation because I am preparing exam in filed of differentiation.
I think that you have to have $A=-\frac{l^2}{2k}$, not $-\frac{-l^2}{2k}$. (note that $k$ is negative)
Let $(s,t)$ be the point on the curve. $$t=\frac{2s-1}{s-1},\qquad s\gt 1\tag1$$
Since $y'=-1/(x-1)^2$, the tangent line is given by $$y-t=-\frac{1}{(s-1)^2}(x-s)$$
Thus, from $(1)$, the area of the triangle is given by $$\begin{align}A&=\frac 12\left(t+\frac{s}{(s-1)^2}\right)(s+t(s-1)^2)\\&=\frac 12\left(\frac{2s-1}{s-1}+\frac{s}{(s-1)^2}\right)\left(s+\frac{2s-1}{s-1}\cdot (s-1)^2\right)\\&=\frac 12\left(\frac{2s^2-2s+1}{s-1}\right)^2\end{align}$$
Let $f(s)=(2s^2-2s+1)/(s-1)$. Then, for $s\gt 1$, $$f'(s)=\frac{2s^2-4s+1}{(s-1)^2}=0\iff s=\frac{2+\sqrt 2}{2}$$
Since $f(s)\gt 0$ for $s\gt 1$, the minimum value of the area is $$\frac 12\left(f\left(\frac{2+\sqrt 2}{2}\right)\right)^2=\frac 12(2+2\sqrt 2)^2=\color{red}{6+4\sqrt 2}$$