## Calculus (3rd Edition)

$$\frac{c}{n}$$
We find the derivative: $$f^{\prime}(x)=nx^{n-1}$$ Then at $x=c, m=f^{\prime}(c)=nc^{n-1}$. Hence, the tangent line is: \begin{aligned} \frac{y-y_{1}}{x-x_{1}} &=m \\ \frac{y-c^n}{x-c} &=nc^{n-1} \\ y &=nc^{n-1} (x-c)+c^n \end{aligned} Since the tangent line intersects with the $x-$axis at $x=0,$ then $Q$ has the coordinates $(c-\frac{c}{n},0), R$ has coordinates $(c,0)$, and the subtangent is $$c-\left(c-\frac{c}{n}\right)=\frac{c}{n}$$