What is the equation of the line that is normal to $f (x) = \frac{e^{2 x - 2}}{\sqrt{2 x - 2}}$ $f(x)= e^(2x-2) /sqrt( 2x-2)$ at $x = 3$ $x=3$ ?

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Answer 1 · 2018-07-04T15:30:54

$y = - \frac{8}{7 \cdot e^{4}} x + \frac{e^{4}}{2} + \frac{24}{7 \cdot e^{4}}$ $y=-8/(7*e^4)x+e^4/2+24/(7*e^4)$

Writing your function in the form

$f (x) = e^{2 x - 2} {(2 x - 2)}^{- \frac{1}{2}}$ $f(x)=e^(2x-2)(2x-2)^(-1/2)$
then we can use the product and the chain rule, doing this we get

$f'(x)=2e^(2x-2)(2x-2)^(-1/2)+e^(2x-2)(2x-2)^(-3/2)*(-1/2)*2$
so we get

$f(3)=e^4-e^4/8=7/8e^4$

now we can find the slope of the normal line

$m_N=-8/(7e^4)$

now we get

$f(3)=e^4/2$
so we get

$y=-8/(7e^4)x+e^4/2+24/(7e^4)$

What is the equation of the line that is normal to f(x)= e^(2x-2) /sqrt( 2x-2) f(x)=e2x−2√2x−2f(x)= e^(2x-2) /sqrt( 2x-2) at x=3 x=3 x=3 ?