The smallest value the expression $x^{2} + 6 x + 8$ attains on the set ${x \in R ∣ ∣ x^{2} - 2 x - 8 \leq 0}$ is

0

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Solution

The correct option is B $0$
We will find set of above equation ;

$x^{2} - 2 x - 8 \leq 0 \Rightarrow (x - 4) (x + 2) \leq 0$

$∴ x ϵ [- 2, 4]$

Roots of the given equation $x^{2} + 6 x + 8 = 0$ are - 2,-4

Differentiating the equation, we get $2 x + 6 > 0$ in the interval [-2,4] $\Rightarrow$ it is an increasing function in that interval

$∴$ minimum value exists at $x = - 2$

At $x = - 2$ , smallest value occurs $= 0$

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Similar questions

{log}_{(x^{2} + 6 x + 8)} ({log}_{(2 x^{2} + 2 x + 3)} (x^{2} - 2 x)) = 0

x^{2} - 6 x + 8 + λ (x^{2} - 4 x + 3) = 0, \forall λ \in R - {- 1}

X^{2} - 6 X + 8 + λ (X^{2} - 4 X + 3) = 0, \forall

λ \in R - {- 1}

x

\frac{x^{2} + 2 x - 3}{x^{2} + 2 x - 8}

x : ∣ ∣ x^{2} + 2 x - 8 ∣ ∣ + x - 2 = 0

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