What is the equation of the normal line of f(x)=xlnx-e^-x at x=2?

Mar 26, 2016

$0.5469 x + y = 2.34477$

Explanation:

As $f \left(x\right) = x \ln x - {e}^{-} x$, at $x = 2$, we have

$f \left(2\right) = 2 \ln 2 - {e}^{-} 2 = 2 \times 0.69315 - 0.13533 = 1.25097$

Hence normal passes through $\left(2 , 1.25097\right)$

As $f \left(x\right) = x \ln x - {e}^{-} x$, $f ' \left(x\right)$ is given by

$\left(1 \times \ln x + x \times \frac{1}{x}\right) - {e}^{- x} \times \left(- 1\right)$ or

$\left(\ln x + 1\right) + {e}^{- x}$ and at $x = 2$, slope of tangent would be

$f ' \left(2\right) = \left(\ln 2 + 1\right) + {e}^{- 2} = 0.69315 + 1 + 0.13533 = 1.82848$

Hence slope of normal would be $- \frac{1}{1.82848} = - 0.5469$

Hence equation of normal would be

$\left(y - 1.25097\right) = - 0.5469 \left(x - 2\right)$ or

$y - 1.25097 = - 0.5469 x + 2 \times 0.5469$ or

$0.5469 x + y = 1.0938 + 1.25097$ or

$0.5469 x + y = 2.34477$