What is the equation of the line that is normal to $f (x) = e^{2 x} \sqrt{2 x + 2}$ $f(x)= e^(2x) sqrt( 2x+2)$ at $x = 1$ $x=1$ ?

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Answer 1 · 2016-05-29T02:03:52

$y - 2 e^{2} = (- \frac{2}{9 e^{2}}) (x - 1)$ $y- 2e^2= (-2/(9e^2))(x-1)$

At x=1, f(x)= $2 e^{2}$ $2e^2$ . The normal is required at point (1, $2 e^{2}$ $2e^2$ )

For slope of tangent to f(x), f'(x) = $2 e^{2 x} \sqrt{2 x + 2} + \frac{e^{2 x}}{\sqrt{2 x + 2}}$ $2e^(2x) sqrt(2x+2)+ e^(2x) / sqrt(2x+2$ .

At x=1, the slope of tangent = $4 e^{2} + \frac{e^{2}}{2}$ $4e^2+e^2 /2$ = $\frac{9 e^{2}}{2}$ $(9e^2) /2$

Slope of normal would thus be $- \frac{2}{9 e^{2}}$ $-2/(9e^2)$

Equation of normal line would be

$y - 2 e^{2} = (- \frac{2}{9 e^{2}}) (x - 1)$ $y- 2e^2= (-2/(9e^2))(x-1)$

What is the equation of the line that is normal to f(x)=e2x√2x+2f(x)= e^(2x) sqrt( 2x+2) at x=1 x=1 ?