Q. The equation of the tangent to the curve $y = 4xe^{x}$at $\left( - 1,\frac{- 4}{e} \right)$is
A) $Y = - 1$
B) $y = - \frac{4}{e}$
C) $x = - 1\$
D) $x = \frac{- 4}{e}$
Explanation
(b)
Given curve is $y = 4xe^{x}$
$\frac{dy}{dx} = 4e^{x} + 4xe^{x}$
At $\left( - 1, - \frac{4}{e} \right),\left( \frac{dy}{dx} \right)_{\left( - 1, - \frac{4}{e} \right)} = 4e^{- 1} + 4( - 1)e^{- 1} = 0$
$\therefore\ \$Equation of tangent is
$\left( y + \frac{4}{e} \right) = 0(x + 1)$
$\Rightarrow \ \ \ \ y = - \frac{4}{e}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.