Respuesta :
Answer:
Step-by-step explanation:
given a point [tex](x_0,y_0)[/tex] the equation of a line with slope m that passes through the given point is
[tex]y-y_0 = m(x-x_0)[/tex] or equivalently
[tex] y = mx+(y_0-mx_0)[/tex].
Recall that a line of the form [tex]y=mx+b [/tex], the y intercept is b and the x intercept is [tex]\frac{-b}{m}[/tex].
So, in our case, the y intercept is [tex](y_0-mx_0)[/tex] and the x intercept is [tex]\frac{mx_0-y_0}{m}[/tex].
In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph [tex](x_0,\frac{1}{x_0})[/tex]. Which means that [tex]y_0=\frac{1}{x_0}[/tex]
The slope of the tangent line is given by the derivative of the function evaluated at [tex]x_0[/tex]. Using the properties of derivatives, we get
[tex]y' = \frac{-1}{x^2}[/tex]. So evaluated at [tex]x_0[/tex] we get [tex] m = \frac{-1}{x_0^2}[/tex]
Replacing the values in our previous findings we get that the y intercept is
[tex](y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}[/tex]
The x intercept is
[tex] \frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0[/tex]
The triangle in consideration has height [tex]\frac{2}{x_0}[/tex] and base [tex]2x_0[/tex]. So the area is
[tex] \frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2[/tex]
So regardless of the point we take on the graph, the area of the triangle is always 2.
