Let f ( x ) = a x 2 + b x + c , and a , b , c ϵ ℜ and a = 0 .
Suppose f ( x ) > 0 for all x ϵ ℜ . Let g ( x ) = f ( x ) + f ′ ( x ) + f ′ ′ ( x ) .
Then:
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Ora, se f ( x ) > 0 então Δ < 0 , isto é, b 2 − 4 a c < 0 .
g ( x ) = f ( x ) + f ′ ( x ) + f ′ ′ ( x ) g ( x ) = ( a x 2 + b x + c ) + ( 2 a x + b ) + ( 2 a ) g ( x ) = a x 2 + ( 2 a + b ) x + ( 2 a + b + c )
Vejamos o que acontece com o discriminante da nova equação obtida,
a x 2 + ( 2 a + b ) x + ( 2 a + b + c ) = 0 Δ = ( 2 a + b ) 2 − 4 ⋅ a ⋅ ( 2 a + b + c ) Δ = − 4 a 2 − 4 a c + b 2 Δ = < 0 − 4 a 2 + < 0 b 2 − 4 a c
Daí, g ( x ) < o , ∀ x ∈ R .
Good job, sir, but i had to translate the portuguese text in english.
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Neatly presented, but the final step is not the correct solution. The discriminant is negative, so either g(x) is always positive or always negative. An additional step is needed to determine which is the case.
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The function g(x) is always a positive, concave-up parabola given that:
1) The discriminant for g(x) is always negative (i.e. roots are complex conjugates), 2) The leading coefficient of g(x) is positive by virtue that f(x) > 0 for all real x.
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Since f ( x ) > 0 ∀ x , we know a > 0 , c > 0 , and b 2 − 4 a c < 0 .
Now g ( x ) = a x 2 + ( 2 a + b ) x + 2 a + b + c which has discriminant Δ = ( 2 a + b ) 2 − 4 a ( 2 a + b + c ) = − 4 a 2 + b 2 − 4 a c < 0 .
Therefore either g ( x ) > 0 or g ( x ) < 0 for all x .
Finally, g ( − 1 ) = a + c > 0 and thus g ( x ) > 0 ∀ x .