Answer to Question #125669 in Calculus for Samuel kassapa

Question

Find the points on the graph of the function f(x)=x+1/x-1 where the slope of the tangent line is equal to -1/3 Determine the equations of the tangent and normal line.

Answer 1

Equation of tangent line is:

$y-y_0=f'(x_0)(x-x_0)$

We have:

$f'(x_0)=-1/3$

Then:

$f'(x)=\frac{x-1-x-1}{(x-1)^2}=-\frac{2}{(x-1)^2}$

$-\frac{2}{(x_0-1)^2}=-1/3$

$(x_0-1)^2=6$

$x_0=\sqrt{6}+1$ or $x_0=-\sqrt{6}-1$

$y_0=\frac{\sqrt{6}+2}{\sqrt{6}}$ or $y_0=\frac{\sqrt{6}}{\sqrt{6}+2}$

So, equations of tangent lines:

$y-\frac{\sqrt{6}+2}{\sqrt{6}}=-\frac{1}{3}(x-\sqrt{6}-1)$

or

$y-\frac{\sqrt{6}}{\sqrt{6}+2}=-\frac{1}{3}(x+\sqrt{6}+1)$

Equation of normal line is:

$y-y_0=-\frac{1}{f'(x_0)}(x-x_0)$

So:

$y-\frac{\sqrt{6}+2}{\sqrt{6}}=3(x-\sqrt{6}-1)$

or

$y-\frac{\sqrt{6}}{\sqrt{6}+2}=3(x+\sqrt{6}+1)$