What is the equation of the line normal to  f(x)=ln(xe^(3x)) at  x=1?

Apr 22, 2018

$y = \left(- \frac{1}{4}\right) x + \frac{13}{4}$

Explanation:

$f \left(x\right) = \ln \left(x {e}^{3 x}\right)$
$f \left(1\right) = \ln \left(1 \cdot {e}^{3}\right) = 3 \ln \left(e\right) = 3$
$f \left(x\right) = \ln \left(x\right) + 3 x \cdot \ln \left(e\right) = \ln \left(x\right) + 3 x$
$f ' \left(x\right) = \frac{1}{x} + 3$
$f ' \left(1\right) = 4$ => slope of the tangent at x = 1:
Slope of the normal at x = 1 is: -1/4:
$m = - \frac{1}{4} , \left({x}_{1} , {y}_{1}\right) = \left(1 , 3\right)$
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - 3 = - \frac{1}{4} \left(x - 1\right)$
$y = \left(- \frac{1}{4}\right) x + \frac{13}{4}$

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