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

The maximum area of a rectangle that can be inscribed in a circle of radius $2 \text{ unit}$ is (in square unit)

The coordinates of the point $P$ on the graph of the function $y = e^{-|x|}$ where the portion of the tangent intercepted between the coordinate axes has the greatest area,is

If $y = a \log x + bx^2 + x$ has extreme values at $x = 1$ and $x = 2$,then $(a, b) = \dots$

In a regular triangular prism,the distance from the centre of one base to one of the vertices of the other base is $l$. Find the altitude of the prism for which the volume is greatest.

The maximum volume (in cubic units) of the cylinder which can be inscribed in a sphere of diameter $6$ units is

Vedclass Test Series

Exam Paper Generator

Online Exam Module