QuadEq

Calculus Level 2

Let f ( x ) = a x 2 + b x + c f(x) = a{ x }^{ 2 } + bx + c , and a , b , c ϵ a, b, c \epsilon \Re and a 0 a \neq 0 .

Suppose f ( x ) > 0 f(x)>0 for all x ϵ x \epsilon \Re . Let g ( x ) = f ( x ) + f ( x ) + f ( x ) g(x) = f(x) + f'(x) + f''(x) .

Then:

g ( x ) < 0 x ϵ g(x) < 0 ∀x \epsilon \Re g ( x ) = 0 x ϵ g(x) = 0 ∀x \epsilon \Re g ( x ) g(x) has real roots g ( x ) > 0 x ϵ g(x) > 0 ∀x \epsilon \Re

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2 solutions

Stuart Price
Feb 17, 2015

Since f ( x ) > 0 x f(x)>0 \,\forall x , we know a > 0 a>0 , c > 0 c>0 , and b 2 4 a c < 0 b^2-4ac<0 .

Now g ( x ) = a x 2 + ( 2 a + b ) x + 2 a + b + c g(x)=ax^2+(2a+b)x+2a+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. \Delta = (2a+b)^2-4a(2a+b+c) = -4a^2+b^2-4ac < 0.

Therefore either g ( x ) > 0 g(x)>0 or g ( x ) < 0 g(x)<0 for all x x .

Finally, g ( 1 ) = a + c > 0 g(-1)=a+c>0 and thus g ( x ) > 0 x g(x)>0\,\forall x .

Daniel Ferreira
Feb 21, 2015

Ora, se f ( x ) > 0 f(x) > 0 então Δ < 0 \Delta < 0 , isto é, b 2 4 a c < 0 b^2 - 4ac < 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 ) g(x) = f(x) + f'(x) + f''(x) \\\\ g(x) = (ax^2 + bx + c) + (2ax + b) + (2a) \\\\ g(x) = ax^2 + (2a + b)x + (2a + 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 = 4 a 2 < 0 + b 2 4 a c < 0 ax^2 + (2a + b)x + (2a + b + c) = 0 \\\\ \Delta = (2a + b)^2 - 4 \cdot a \cdot (2a + b + c) \\\\ \Delta = - 4a^2 - 4ac + b^2 \\\\ \Delta = \underbrace{-4a^2}_{<0} + \underbrace{b^2 - 4ac}_{< 0}

Daí, g ( x ) < o , x R \boxed{g(x) < o}, \forall x \in \mathbb{R} .

Good job, sir, but i had to translate the portuguese text in english.

Neeraj Snappy - 6 years, 3 months ago

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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.

Stuart Price - 6 years, 3 months ago

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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.

tom engelsman - 4 years, 6 months ago

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