The function f(x) = ax + frac{b}{x} where a, | vedclass

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

Question

(D) Given the function $f(x) = ax + \frac{b}{x}$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f'(x) = a - \frac{b}{x^2}$.S

Vedclass · Accepted Answer

$\sqrt{\frac{b}{a}}$

Vedclass · Answer

(D) Given the function $f(x) = ax + \frac{b}{x}$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f'(x) = a - \frac{b}{x^2}$.Setting $f'(x) = 0$ for critical points:$a - \frac{b}{x^2} = 0 \Rightarrow a = \frac{b}{x^2} \Rightarrow x^2 = \frac{b}{a} \Rightarrow x = \sqrt{\frac{b}{a}}$ (since $x > 0$).Now,we find the second derivative to check for minima:$f''(x) = \frac{d}{dx}(a - bx^{-2}) = 0 - b(-2)x^{-3} = \frac{2b}{x^3}$.Since $b > 0$ and $x > 0$,$f''(x) = \frac{2b}{x^3} > 0$ for all $x > 0$.Since $f''(x) > 0$ at $x = \sqrt{\frac{b}{a}}$,the function attains its least value at $x = \sqrt{\frac{b}{a}}$.

The function $f(x) = ax + \frac{b}{x}$ where $a, b, x > 0$ takes on the least value at $x$ equal to:

Similar Questions

The height of a right circular cone of maximum volume inscribed in a sphere of diameter $a$ is

Let a function $f(x)$ be defined as $f(x) = \begin{cases} \cos^{-1}(\mu) + x^2, & 0 < x < 1 \\ 4x, & x \geqslant 1 \end{cases}$. The function $f(x)$ can have a local minimum at $x = 1$ if the value of $\mu$ lies in the interval:

Find the point where $f(x) = 2\sin x + \cos 2x$ has a maximum value in the interval $0 \leq x \leq 2\pi$.

If $x_0$ is the point of local minima of $f(x) = \bar{a} \cdot (\bar{b} \times \bar{c})$ where $\bar{a} = x \hat{i} - 2 \hat{j} + 3 \hat{k}$,$\bar{b} = -2 \hat{i} + x \hat{j} - \hat{k}$,$\bar{c} = 7 \hat{i} - 2 \hat{j} + x \hat{k}$,then the value of $\bar{a} \cdot \bar{b}$ at $x = x_0$ is

If $xy = c^2$,then the minimum value of $ax + by$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module