The minimum value of f x = x + 4 x + 2 is, - 1, - 2, 1, 2

f x = x + 4 x + 2 For minima, f ' x = 0 1 - 4 x + 2 2 = 0 x + 2 2 = 4 ⇒ x + 2 = ± 2 ⇒ x = 0 , - 4 f " x = 8 x + 2 - 3 At x = 0 , f x will have minima. i.e. the minimum value f 0 = 2

The minimum value of f x = x + 4 x + 2 is

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The minimum value of

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

MathematicsApplication of DerivativesAP EAMCETAP EAMCET 2022 (04 Jul Shift 1)

Options:

A $- 1$
B $- 2$
C $1$
D $2$

Solution:

2997 Upvotes Verified Answer

The correct answer is:

2

$f (x) = x + \frac{4}{x + 2}$
For minima, $f^{'} (x) = 0$
$1 - \frac{4}{{(x + 2)}^{2}} = 0$
${(x + 2)}^{2} = 4 \Rightarrow x + 2 = \pm 2 \Rightarrow x = 0, - 4$
$f^{"} (x) = 8 {(x + 2)}^{- 3}$
At $x = 0, f (x)$ will have minima.
i.e. the minimum value $f (0) = 2$

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