Let f(x) = x^3 e^{-3x}, x 0. Then the maxim | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the function $f(x) = x^3 e^{-3x}$ for $x > 0$. To find the maximum value,we first find the derivative $f'(x)$ using the product rule: $f'(x)

Vedclass · Accepted Answer

$e^{-3}$

Vedclass · Answer

(A) Given the function $f(x) = x^3 e^{-3x}$ for $x > 0$. To find the maximum value,we first find the derivative $f'(x)$ using the product rule: $f'(x) = \frac{d}{dx}(x^3) \cdot e^{-3x} + x^3 \cdot \frac{d}{dx}(e^{-3x})$ $f'(x) = 3x^2 e^{-3x} + x^3 (-3 e^{-3x})$ $f'(x) = 3x^2 e^{-3x} (1 - x)$ Setting $f'(x) = 0$ for critical points: $3x^2 e^{-3x} (1 - x) = 0$ Since $x > 0$ and $e^{-3x} 
eq 0$,we have $1 - x = 0$,which gives $x = 1$. To confirm it is a maximum,we check the sign of $f'(x)$ around $x = 1$. For $x  0$ and for $x > 1$,$f'(x) Thus,$f(x)$ has a local maximum at $x = 1$. The maximum value is $f(1) = (1)^3 e^{-3(1)} = e^{-3}$.

Let $f(x) = x^3 e^{-3x}, x > 0$. Then the maximum value of $f(x)$ is

Similar Questions

Let $AP$ and $BQ$ be two vertical poles at points $A$ and $B$,respectively. If $AP=16 \, m, BQ=22 \, m$ and $AB=20 \, m,$ then find the distance of a point $R$ on $AB$ from the point $A$ such that $RP^2 + RQ^2$ is minimum. (in $, m$)

The condition that $f(x) = ax^3 + bx^2 + cx + d$ has no extreme value is

The function $f(x)=2|x|+|x+2|-||x+2|-2|x||$ has a local minimum or a local maximum at $x=$

Let $f: R \to R$ be defined by $f(x) = \begin{cases} k - 2x, & \text{if } x \leqslant -1 \\ 2x + 3, & \text{if } x > -1 \end{cases}$. If $f$ has a local minimum at $x = -1$,then what is the possible value of $k$?

If $f(x)=3 \sqrt[3]{x^2}-x^2$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module