The minimum value of the function f(x) = x lo | vedclass

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

Question

(D) Given the function $f(x) = x \log x$. To find the minimum value,we first find the derivative $f'(x)$. Using the product rule: $f'(x) = x \cdot \fr

Vedclass · Accepted Answer

$-\frac{1}{e}$

Vedclass · Answer

(D) Given the function $f(x) = x \log x$. To find the minimum value,we first find the derivative $f'(x)$. Using the product rule: $f'(x) = x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(x) = x \cdot \frac{1}{x} + \log x = 1 + \log x$. For critical points,set $f'(x) = 0$: $1 + \log x = 0 \implies \log x = -1 \implies x = e^{-1} = \frac{1}{e}$. Now,find the second derivative $f''(x)$: $f''(x) = \frac{d}{dx}(1 + \log x) = \frac{1}{x}$. Evaluate $f''(x)$ at $x = \frac{1}{e}$: $f''(\frac{1}{e}) = \frac{1}{1/e} = e$. Since $e > 0$,the function has a local minimum at $x = \frac{1}{e}$. The minimum value is $f(\frac{1}{e}) = \frac{1}{e} \log(\frac{1}{e}) = \frac{1}{e} \log(e^{-1}) = \frac{1}{e} (-1) = -\frac{1}{e}$.

The minimum value of the function $f(x) = x \log x$ is

Similar Questions

Maximum value of $x(1 - x)^2$ when $0 \le x \le 2$,is

The abscissae of the points of the curve $y = x(x - 2)(x - 4)$ where the tangents are parallel to the $x$-axis are obtained as:

If $p(x)$ is a polynomial of degree three that has a local maximum value $8$ at $x=1$ and a local minimum value $4$ at $x=2$,then $p(0)$ is equal to:

If the area of a right-angled triangle with hypotenuse $5$ is maximum,then its perimeter is

At $x=0$,$f(x)=\cos x-1+\frac{x^2}{2}-\frac{x^3}{3}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module