# What is the equation of the line that is normal to f(x)=(2x-1)/e^x at  x=1/2 ?

Equation of the normal line

$\textcolor{b l u e}{y = - \frac{1}{2} {e}^{\frac{1}{2}} \left(x - \frac{1}{2}\right)}$

#### Explanation:

$f \left(x\right) = \frac{2 x - 1}{e} ^ x$ at $x = \frac{1}{2}$

Solve for the point $\left({x}_{1} , {y}_{1}\right)$

let ${x}_{1} = \frac{1}{2}$

${y}_{1} = f \left({x}_{1}\right) = \frac{2 {x}_{1} - 1}{e} ^ \left({x}_{1}\right)$

${y}_{1} = f \left(\frac{1}{2}\right) = \frac{2 \left(\frac{1}{2}\right) - 1}{e} ^ \left(\frac{1}{2}\right)$

${y}_{1} = f \left(\frac{1}{2}\right) = \frac{1 - 1}{e} ^ \left(\frac{1}{2}\right) = 0$

${y}_{1} = 0$

Our point $\left({x}_{1} , {y}_{1}\right) = \left(\frac{1}{2} , 0\right)$

Solve for the slope $m$

$f \left(x\right) = \frac{2 x - 1}{e} ^ x$

$f ' \left(x\right) = \frac{{e}^{x} \cdot \frac{d}{\mathrm{dx}} \left(2 x - 1\right) - \left(2 x - 1\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right)}{{e}^{x}} ^ 2$

$f ' \left(x\right) = \frac{\left({e}^{x} \cdot 2\right) - \left(2 x - 1\right) \cdot \left({e}^{x}\right)}{{e}^{x}} ^ 2$

$f ' \left(x\right) = \frac{\left(2 {e}^{x} - 2 x {e}^{x} + {e}^{x}\right)}{{e}^{x}} ^ 2$

Slope $m = f ' \left(\frac{1}{2}\right) = \frac{\left(2 {e}^{\frac{1}{2}} - 2 \left(\frac{1}{2}\right) {e}^{\frac{1}{2}} + {e}^{\frac{1}{2}}\right)}{{e}^{\frac{1}{2}}} ^ 2$

$m = f ' \left(\frac{1}{2}\right) = \frac{\left(2 {e}^{\frac{1}{2}} - {e}^{\frac{1}{2}} + {e}^{\frac{1}{2}}\right)}{e}$

$m = f ' \left(\frac{1}{2}\right) = \left(2 {e}^{- \frac{1}{2}}\right)$

For the perpendicular line we need

m_p=-1/m=-1/((2e^(-1/2))

${m}_{p} = - \frac{1}{2} {e}^{\frac{1}{2}}$

Solve for the equation of the line using $\textcolor{b l u e}{\text{Point-Slope Form}}$

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

$y - 0 = - \frac{1}{2} {e}^{\frac{1}{2}} \left(x - \frac{1}{2}\right)$

$\textcolor{red}{y = - \frac{1}{2} {e}^{\frac{1}{2}} \left(x - \frac{1}{2}\right)}$

Kindly see the graph of $f \left(x\right) = \frac{2 x - 1}{e} ^ x$ (colored red) and the normal line $y = - \frac{1}{2} {e}^{\frac{1}{2}} \left(x - \frac{1}{2}\right)$ (colored blue)

God bless....I hope the explanation is useful.