bonjour,
1)
0≤x≤6
2)
f(x) = aire figure
f(6)= aire carré
AM² = 6²
f(6)=36
f(0)= aire triangle =AB²/2
=6² /2
f(0)=18
3)
MB= AB-AM
MB=6-x
4)
x²+(6-x)²/2
5)
=x²+(36-12x+x²)/2
=x²+18-6x+x²/2
=(3/2) x² -6x +18
6)
3/2 ((x-2)² +8)
on développe
3/2(x²-4x+4 +8)
=3/2 x² -12/ 2x +12/2 +24/2
=(3/2) x² -6x +18
= f(x)
forme canonique de f(x)
=a(x-α)²+β
=(3/2) (x-2)² +12
minimum de la fonction
α=2
β=12
soit le sommet (2;12)
f(x) admet en minimum en x = 2 qui vaut
12
