Question:

The global minimum of \(x^3 e^{-|x|}\) for \(x \in (-\infty, \infty)\) occurs at \(x =\) _________ (round off to one decimal place).

Updated On: Nov 25, 2025
Hide Solution
collegedunia
Verified By Collegedunia

Correct Answer: -3

Solution and Explanation

To find the global minimum of \(f(x) = x^3 e^{-|x|}\), we differentiate \(f(x)\) in two cases, for \(x \geq 0\) and \(x < 0\).
For \(x \geq 0\), the function is \(f(x) = x^3 e^{-x}\), and for \(x < 0\), the function is \(f(x) = -x^3 e^{x}\). 
The derivative of \(f(x)\) is calculated, and by setting it equal to zero, we find that the global minimum occurs at \( \boxed{-3.0} \). 
 

Was this answer helpful?
0
0

Top Questions on Mean value theorems

Questions Asked in GATE IN exam

View More Questions