What is the equation of the line normal to $f (x) = e^{\sqrt{x} - x}$ $f(x)=e^(sqrtx-x)$ at $x = 4$ $x=4$ ?

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Answer 1 · 2016-09-14T20:24:39

I get $y = \frac{4 e^{2}}{3} x - \frac{16 e^{2}}{3} + \frac{1}{e^{2}}$ $y = (4e^2)/3x-(16e^2)/3+1/e^2$

$f (x) = e^{\sqrt{x} - x}$ $f(x) = e^(sqrtx-x)$

$f (4) = e^{- 2} = \frac{1}{e^{2}}$ $f(4) = e^-2 = 1/e^2$

$f'(x) = e^(sqrtx-x)(1/(2sqrtx)-1)$

$f'(4) =e^(sqrt4-4)(1/(2sqrt4)-1) = e^(2-4)(1/4-1) = e^(-2)(-3/4) = -3/(4e^2)$

The normal line it perpendicular to the tangent, so the slope of the normal line is $-1/(f'(4)) = (4e^2)/3$ .

The line through $(4,1/e^2 )$ with slope $m=(4e^2)/3$ is

$y-1/e^2 = (4e^2)/3(x-4)$ or

$y = (4e^2)/3x-(16e^2)/3+1/e^2$

What is the equation of the line normal to f(x)=e^(sqrtx-x)f(x)=e√x−x f(x)=e^(sqrtx-x) at x=4x=4 x=4?