Fonction

F(1) = -1/2 x1² +1+1/2 = -1/2+1+1/2=1 -1/2(x-1)² + f(1) = -1/2 (x²-2x+1) + 1 (identité remarquable (a-b)² = a²-2ab+b²) = -1/2x²+x-1/2+1 = -1/2x²+x+1/2 = f(x) f(x) = 1 <=> -1/2(x-1)² + f(1) = 1 <=> -1/2 (x-1)² +1 = 1 <=> -1/2 (x-1)² =0 <=> (x-1)² =0 <=> x-1 =0 donc x =1 f(x) = 5 <=> -1/2(x-1)² +f(1) = 5 <=> -1/2(x-1)² +1 =5 <=> -1/2(x-1)² = 4 <=> (x-1)² = 4 x (-2) = -8 Solution impossible car un carré est toujours positif.