Question

1) f: R→R f(x) = x²-4

2) f: R→R f(x) = -x²+3x-2​

Answer 1

Explicație pas cu pas:

1) f: R→R f(x) = x²- 4

$x^{2} - 4 = 0 = > (x - 2)(x + 2) = 0$

$\frac{ - b}{2a} = 0$

$Δ = 0 - 4( - 4) = 16$

punct de minim:

$V(0 ;-\frac{Δ}{4}) = V(0 ;-4)$

$Imf = [-4; + \infty )$

2) f: R→R f(x) = -x² + 3x - 2

$-x² + 3x - 2 = 0 \\ - (x - 1)(x - 2) = 0$

$- \frac{b}{2a} = - \frac{3}{2( - 1)} = \frac{3}{2}$

$Δ = 9 - 4( - 2)( - 1) = 9 - 8 = 1$

punct de maxim:

$V( - \frac{b}{2a} ;-\frac{Δ}{4a}) = V( \frac{3}{2} ; \frac{1}{4} )$

$Imf = ( - \infty ; \frac{1}{4} ]$