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Irina-Kira [14]
3 years ago
12

Prove that the inequality x(x+ 1)(x+ 2)(x+ 3) ≥ −1 holds for all real numbers x.

Mathematics
1 answer:
vredina [299]3 years ago
6 0
(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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Answer:

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Step-by-step explanation:

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Hint: | Distribute 2 over a + 5.

2 (a + 5) = 2 a + 10:

(2 a + 10) - 1 = 3

Hint: | Group like terms in 2 a - 1 + 10.

Grouping like terms, 2 a - 1 + 10 = 2 a + (10 - 1):

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Hint: | Evaluate 10 - 1.

10 - 1 = 9:

2 a + 9 = 3

Hint: | Isolate terms with a to the left hand side.

Subtract 9 from both sides:

2 a + (9 - 9) = 3 - 9

Hint: | Look for the difference of two identical terms.

9 - 9 = 0:

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Hint: | Evaluate 3 - 9.

3 - 9 = -6:

2 a = -6

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of 2 a = -6 by 2:

(2 a)/2 = (-6)/2

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Hint: | Reduce (-6)/2 to lowest terms. Start by finding the GCD of -6 and 2.

The gcd of -6 and 2 is 2, so (-6)/2 = (2 (-3))/(2×1) = 2/2×-3 = -3:

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