"What is the equation of the normal line..." was answered by our experts

Kali Thomas

Answered question

2022-04-09

What is the equation of the normal line of

f (x) = x e^{x}

at x=3?

prangijahnot

Beginner2022-04-10Added 17 answers

Substituting x=3 in the given function, we get y-coordinate of point is obtained as follows

y = f (3) = 3 e^{3}

Differentiating given function w.r.t. x, the slope of tangent

\frac{d y}{d x}

\frac{d y}{d x} = f^{'} (x)

= \frac{d}{d x} (x e^{x})

= x e^{x} + e^{x}

hence the slope of tangent at x=3 is
=f'(3)

= 3 e^{3} + e^{3}

= 4 e^{3}

hence the slope m of normal at the same point x=3 is given as

m = - \frac{1}{f^{'} (3)}

= - \frac{1}{4 e^{3}}

Now, the equation of normal at the point

(3, 3 e^{3})

& having slope

m = - \frac{1}{4 e^{3}}

is
given as follows

y - 3 e^{3} = - \frac{1}{4 e^{3}} (x - 3)

Do you have a similar question?

Recalculate according to your conditions!