Find all values of x at which the tangent line to the given curve passes through the origin for y = 1/(x + 4)
The tangent is a line of slope "m".
The derivative of the function gives this slope.
As the tangent passes through the origin, it's equation is y=mx, as c=0.
Hence, the point of intersection of this tangent and the curve gives us x.
$\displaystyle f'(x)=\frac{(x+4)(0)-1(1)}{(x+4)^2}$
The equation of the tangent(s) through (0,0) is $\displaystyle y=mx=\frac{-x}{(x+4)^2}$
Solving f(x) for the curve = mx for the line finds x.
$\displaystyle \frac{1}{x+4}=\frac{-x}{(x+4)^2}$
Solving this, bearing in mind that x=-4 is not part of the domain of f(x) or f'(x) discovers x.
Hence multiply both fractions by (x+4) before solving for x.
Lines look like this: $\displaystyle (y-y_{0}) = m(x-x_{0})$
We know we need the Origin: $\displaystyle y_{0} = m \cdot x_{0}$
The derivative gives the slope: $\displaystyle m = -\frac{1}{(x_{0}+4)^{2}}$
We know: $\displaystyle y_{0} = \frac{1}{x_{0}+4}$
I'm a little surprised to find only one solution. I expected two solutions when I started the problem. I was wrong.
Let's see what you get.