What is the equation of the line normal to $f(x)=xe^x$ at $x=7$ ?

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Answer 1 · 2016-02-02T14:59:45

Equation of normal line

$y = (7(8e^14 + 1) - x)/(8e^{7})$

$f(7) = 7e^7$

The normal line has to pass through the point $(7,f(7))$ .

$f'(x) = (x+1)e^x$

$f'(7) = 8e^7$

Since the normal line is perpendicular to the tangent line, we have the gradient of the normal line, $m$ , given by

$m = frac{-1}{f'(7)} = -frac{1}{8}e^{-7}$ .

The $y$ -intercept of the line is given by

$c = y - mx$

$= 7e^7 - (-frac{1}{8}e^{-7})(7)$

$= 7e^{-7}(e^14 + 1/8)$

Equation of normal line in slope-intercept form

$y = -frac{1}{8}e^{-7}x + 7e^{-7}(e^14 + 1/8)$

What is the equation of the line normal to f(x)=xe^xf(x)=xe^x at x=7x=7?