Question

Si x ∈ [-2 ; 3 ] ; calcule $\frac{a}{2}$ +4b
Si ( 4+3x-$x^{2}$ ) ∈ [ a,b ]

Answer 1

Explicación paso a paso:

A= $-x^{2}$ +3x + 4 => A= - [ $x^{2}$ - 2( x ) $\frac{3}{2}$ + $\frac{9}{4}$ - $\frac{9}{4}$ ] + 4

A= - [ (x - $\frac{3}{2}$ $)^{2}$ - $\frac{9}{4}$ ] +4

A= - (x - $\frac{3}{2}$ $)^{2}$ + $\frac{9}{4}$ + 4 => A= (x - $\frac{3}{2}$ $)^{2}$ + $\frac{25}{4}$

Del dato formado:

A= - 2 - $\frac{3}{2}$ ≤ x - $\frac{3}{2}$ ≤ 3 - $\frac{3\\}{2}$

A= - $\frac{7}{2}$ ≤ x - $\frac{3}{2}$ ≤ $\frac{3}{2}$

A= 0 (-1) ≤ (-1) (x - $\frac{3}{2}$ $)^{2}$ ≤ $\frac{49}{4}$ (-1)  

A= - $\frac{49}{4}$ + $\frac{25}{4}$ ≤ - (x - $\frac{3}{2}$ $)^{2}$ + $\frac{25}{4}$ ≤ 0 + $\frac{25}{4}$

A= $\frac{-49+25}{4}$ ≤ A ≤ $\frac{25}{4}$

A= - 6 ≤ A ≤ $\frac{25}{4}$

A∈ [ -6 ; $\frac{25}{4}$ ] = [ a,b]

a= -6 ∧ b= $\frac{25}{4}$

∴ $\frac{a}{2}$ + 2b = $\frac{-6}{2}$ +4 . $\frac{25}{4}$

= -3 +25

= 22