The minimum value of f(x) = x + frac{4}{x + 2 | vedclass

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

Question

(D) Given function: $f(x) = x + \frac{4}{x + 2}$ for $x > -2$.To find the critical points,we calculate the derivative $f'(x)$ and set it to $0$:$f'(x)

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given function: $f(x) = x + \frac{4}{x + 2}$ for $x > -2$.To find the critical points,we calculate the derivative $f'(x)$ and set it to $0$:$f'(x) = 1 - \frac{4}{(x + 2)^2}$Setting $f'(x) = 0$:$1 = \frac{4}{(x + 2)^2} \implies (x + 2)^2 = 4$Since $x > -2$,we have $x + 2 = 2$,which gives $x = 0$.Now,we check the second derivative $f''(x) = \frac{8}{(x + 2)^3}$.At $x = 0$,$f''(0) = \frac{8}{2^3} = 1 > 0$.Since $f''(0) > 0$,the function has a local minimum at $x = 0$.The minimum value is $f(0) = 0 + \frac{4}{0 + 2} = \frac{4}{2} = 2$.

The minimum value of $f(x) = x + \frac{4}{x + 2}$ for $x > -2$ is

Similar Questions

The function $f(x) = x^2 \log x$ in the interval $(1, e)$ has

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

Find both the maximum value and the minimum value of $3x^{4}-8x^{3}+12x^{2}-48x+25$ on the interval $[0,3]$.

The point $(0, 5)$ is closest to the curve ${x^2} = 2y$ at

The maximum value of ${\left( {\frac{1}{x}} \right)^x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module