Example 2. Find the point $c$ mentioned in Lagrange's theorem for the function $y=x^{2}+6 x+1$ on the interval $[-1 ; 3]$.
Solution. We have $y(-1)=-4, y(3)=28, A(-1,-4), B(3,28)$, $k_{A B}=\frac{28+4}{3+1}=8$. We need to solve the equation $y^{\prime}=2 x+6=8$. We find $x=1$, i.e., $C$ has coordinates $(1,8)$. The tangent at this point is parallel to the chord $A B$.

## Exercises

Check whether Rolle's theorem is applicable to the given functions on the given intervals, and find the corresponding values of $c$ (if they exist)

1. $f(x)=|x-1|,[0 ; 3]$. 2. $f(x)=\sqrt[3]{x+1},[-2 ; 0]$.
2. $f(x)=x^{2},[-1 ; 2]$.

Find the point $c$ mentioned in Lagrange's theorem for the given functions on the given intervals.
4. $f(x)=|x|-3,[-2 ; 2]$. 5. $f(x)=x^{3}, x \in[-1 ; 1]$.
6. $f(x)=x^{2}-6 x+1,[0 ; 1]$.

## Answers

1.-4. No. 5. $\pm \frac{1}{\sqrt{3}}$. 6. $\frac{1}{2}$.

## § 9. Applications of the Derivative

## 9.1. Equations of the Tangent and Normal to a Curve

In § 2, the equations of the tangent $(t): y=y_{0}+k\left(x-x_{0}\right)$, $k=f^{\prime}\left(x_{0}\right)$ and the normal $(n): y=y_{0}-\frac{1}{k}\left(x-x_{0}\right), k=f^{\prime}\left(x_{0}\right)$ to the curve $y=f(x)$ at the point $A\left(x_{0}, y_{0}\right), y_{0}=f\left(x_{0}\right)$ are provided.

If the curve is given implicitly or parametrically, the slope $k=f^{\prime}\left(x_{0}\right)$ must be found as the derivative of the implicitly or parametrically defined function.

## Examples with Solutions