The minimum value of frac{(5 + x)(2 + x)}{1 + | vedclass

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

Question

(C) Let $f(x) = \frac{(5 + x)(2 + x)}{1 + x}$.We can rewrite the expression as:$f(x) = \frac{x^2 + 7x + 10}{x + 1} = \frac{x(x + 1) + 6(x + 1) + 4}{x

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let $f(x) = \frac{(5 + x)(2 + x)}{1 + x}$.We can rewrite the expression as:$f(x) = \frac{x^2 + 7x + 10}{x + 1} = \frac{x(x + 1) + 6(x + 1) + 4}{x + 1} = x + 6 + \frac{4}{x + 1}$.To find the minimum,we differentiate $f(x)$ with respect to $x$:$f'(x) = 1 - \frac{4}{(x + 1)^2}$.Setting $f'(x) = 0$ for critical points:$1 = \frac{4}{(x + 1)^2} \implies (x + 1)^2 = 4 \implies x + 1 = \pm 2$.Since $x$ is non-negative,$x + 1 = 2$,which gives $x = 1$.Using the second derivative test:$f''(x) = \frac{8}{(x + 1)^3}$.At $x = 1$,$f''(1) = \frac{8}{8} = 1 > 0$,so $x = 1$ is a point of local minima.The minimum value is $f(1) = \frac{(5 + 1)(2 + 1)}{1 + 1} = \frac{6 	imes 3}{2} = 9$.

The minimum value of $\frac{(5 + x)(2 + x)}{1 + x}$ for non-negative real $x$ is

Similar Questions

Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

If $f(x) = x + \frac{1}{x}$,$x \neq 0$,then the local maximum and minimum values of the function $f$ are respectively....

If $x + y = 8$,find the maximum value of $xy$.

Divide $10$ into two parts such that the sum of twice the first part and the square of the second part is minimum. The two parts are:

Statement-$I$: Let the function $f(x) = \begin{cases} -\frac{x}{2} & x < 0 \\ 7x + 8 & x \geq 0 \end{cases}$. Then $f(x)$ has a local minimum at $x = 0$.
Statement-$II$: If $f(a) < f(a - h)$ and $f(a) < f(a + h)$ for a sufficiently small $h > 0$,then $f(x)$ has a local minimum at $x = a$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module