Prove that the inequality
x(x+ 1)(x+ 2)(x+ 3) ≥ −1
holds for all real numbers x.
1 answer:
(x²+x)(x²+5x+6) = x^4 + 6x³ + 11x² + 6x + 1 ≥0
f(x) = x^4 + 6x³ + 11x² + 6x + 1
f'(x) = 4x³ + 18x² +22x +6
racines x1 = -0,301 x2 = -1,5 x3 = -2,618
variations
x -2,62 -1,5 -0,3
f'(x) - 0 + 0 - 0 +
f(x) -inf \ 0 / \ 0 /
on voit que cette fonction st toujours positive
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