Find the area of a triangle enclosed by axes and tangent to y = 1/x
Well, let's see. If the function is $\displaystyle \displaystyle \begin{align*} y = \frac{1}{x} \end{align*}$, the derivative is $\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} = -\frac{1}{x^2} \end{align*}$. This tells us the gradient of the tangent line at any point x.
So the tangent line is of the form $\displaystyle \displaystyle \begin{align*} y = \left(-\frac{1}{x^2}\right)x + c = -\frac{1}{x} + c \end{align*}$.
c is the y intercept. To find the x intercept, we'll let y = 0, and we find that
$\displaystyle \displaystyle \begin{align*} 0 &= -\frac{1}{x} + c \\ \frac{1}{x} &= c \\ x &= \frac{1}{c} \end{align*}$.
So the height of the triangle is c, the length is $\displaystyle \displaystyle \begin{align*} \frac{1}{c} \end{align*}$, which means the area is
$\displaystyle \displaystyle \begin{align*} A &= \frac{1}{2}(c)\left(\frac{1}{c}\right) \\ &= \frac{1}{2} \end{align*}$
So the area is $\displaystyle \displaystyle \begin{align*} \frac{1}{2}\,\textrm{unit}^2 \end{align*}$
Don't think that is correct. Example consider tangent at (1,1) Gradient is -1 Equation of tangent is y=-x+2 Intercepts are both 2 Area of triangle is 2
In general consider tangent at (k,m) Gradient of tangent is -1/k^2 Get the equation of the tangent and the intercepts. Knowing also that km=1 wil get area of triangle=2