Question

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

Answer 1

`y- 2e^2= (-2/(9e^2))(x-1)`

Explanation:

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

For slope of tangent to f(x), f'(x) = `2e^(2x) sqrt(2x+2)+ e^(2x) / sqrt(2x+2` .

At x=1, the slope of tangent =`4e^2+e^2 /2` = `(9e^2) /2`

Slope of normal would thus be `-2/(9e^2)`

Equation of normal line would be

`y- 2e^2= (-2/(9e^2))(x-1)`