# What is the equation of the normal line of f(x)=3^x-x at x=-3?

$y - \frac{82}{27} = \frac{27}{27 - \setminus \ln 3} \left(x + 3\right)$

#### Explanation:

Substituting $x = - 3$ in the given function: $f \left(x\right) = {3}^{x} - x$ to get y-coordinate of given point as follows

$y = f \left(- 3\right)$

$= {3}^{- 3} - \left(- 3\right)$

$= \frac{82}{27}$

Now, the slope of tangent $\frac{\mathrm{dy}}{\mathrm{dx}}$ at any point to the given curve: $y = {3}^{x} - x$ is given by differentiating given function w.r.t. $x$ as follows

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right)$

$= \frac{d}{\mathrm{dx}} \left({3}^{x} - x\right)$

$= {3}^{x} \setminus \ln 3 - 1$

hence the slope of tangent at $\left(- 3 , \frac{82}{27}\right)$

$f ' \left(- 3\right) = {3}^{- 3} \setminus \ln 3 - 1$

$= \frac{\setminus \ln 3 - 27}{27}$

hence the slope $\left(m\right)$ of normal at the same point $\left(- 3 , \frac{82}{27}\right)$

$m = - \frac{1}{f ' \left(- 3\right)}$

$= - \frac{1}{\frac{\setminus \ln 3 - 27}{27}}$

$= \frac{27}{27 - \setminus \ln 3}$

hence the equation of normal at the point $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(- 3 , \frac{82}{27}\right)$ & having slope $m = \frac{27}{27 - \setminus \ln 3}$ is given by following formula

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - \frac{82}{27} = \frac{27}{27 - \setminus \ln 3} \left(x - \left(- 3\right)\right)$

$y - \frac{82}{27} = \frac{27}{27 - \setminus \ln 3} \left(x + 3\right)$