The minimum value of fx-x-4x-2- x-0 is

F(x)=x+4x+2 ⇒ f'(x)=1 4(x+2)^2 ⇒ f'(x)=x(x+4)(x+2)^2≥0, ∀ x gt;0 So f(x) is increasing function ⇒ f(x)≥ f(0) So, x=0 is the only critical point f(0)=2 when x→ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Condition for Strictly Increasing

The minimum v...

Question

The minimum value of $f (x) = x + \frac{4}{x + 2}, x \geq 0$ is

- 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $2$
$f (x) = x + \frac{4}{x + 2}$
$\Rightarrow f^{'} (x) = 1 - \frac{4}{(x + 2)^{2}}$
$\Rightarrow f^{'} (x) = \frac{x (x + 4)}{(x + 2)^{2}} \geq 0, \forall x > 0$
So $f (x)$ is increasing function
$\Rightarrow f (x) \geq f (0)$
So, $x = 0$ is the only critical point
$f (0) = 2$
when $x \to \infty, f \to \infty$
$∴$ The minimum value of $f (x)$ is $2$

Suggest Corrections

Similar questions

Q. The minimum value of

f (x) = x + \frac{4}{x + 2}

Q. The absolute minimum value of

f (x) = \frac{x - 2}{\sqrt{x^{2} + 1}}

f (x) = \frac{(x - 2) (x - 1)}{(x - 3)} \forall x > 3

. The minimum value of f(x) is equal to

Q. The minimum value of

f (x) = | 3 - x | + | 2 + x | + | 5 - x |

Q. The minimum value of

f (x) = | x - 1 | + | x - 2 | + | x - 3 |

is:

Join BYJU'S Learning Program

Explore more

Condition for Strictly Increasing

Standard XII Mathematics

Join BYJU'S Learning Program

The minimum value of f(x)=x+4x+2, x≥0 is

The correct option is D 2 f(x)=x+4x+2 ⇒f′(x)=1−4(x+2)2 ⇒f′(x)=x(x+4)(x+2)2≥0,∀x>0 So f(x) is increasing function ⇒f(x)≥f(0) So, x=0 is the only critical point f(0)=2 when x→∞, f→∞ ∴ The minimum value of f(x) is 2

The minimum value of $f (x) = x + \frac{4}{x + 2}, x \geq 0$ is