max ⁡ min ⁡ − min ⁡ max ⁡ \max_{\min}-\min_{\max}

Algebra Level pending

Given two functions

{ f ( x ) = x 2 − 2 ( a + 2 ) x + a 2 g ( x ) = − x 2 + 2 ( a − 2 ) x − a 2 + 8 \begin{cases} f(x)=x^2-2(a+2)x+a^2 \\ g(x)=-x^2+2(a-2)x-a^2+8\end{cases}

Let H 1 ( x ) = max ⁡ { f ( x ) , g ( x ) } H_1(x)=\max\{f(x),g(x)\} , where max ⁡ { f ( x ) , g ( x ) } \max\{f(x),g(x)\} denote the bigger value of f ( x ) , g ( x ) f(x),g(x) .

H 2 ( x ) = min ⁡ { f ( x ) , g ( x ) } H_2(x)=\min\{f(x),g(x)\} , where min ⁡ { f ( x ) , g ( x ) } \min\{f(x),g(x)\} denote the smaller value of f ( x ) , g ( x ) f(x),g(x) .

Then H 1 ( x ) H_1(x) has the minimum value A A and H 2 ( x ) H_2(x) has the maximum value B B .

For all a ∈ R a \in \mathbb R , find the value of A − B A-B .

− 16 -16 a 2 + 2 a − 16 a^2+2a-16 16 16 a 2 − 2 a − 16 a^2-2a-16

Alice Smith by Alice Smith , 20, China

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1 solution

We have f ( x ) = x 2 − 2 ( a + 2 ) x + a 2 ≥ − 4 ( a + 1 ) f(x)=x^2-2(a+2)x+a^2\geq -4(a+1) and g ( x ) = − x 2 + 2 ( a − 2 ) x − a 2 + 8 ≤ 4 ( 3 − a ) g(x)=-x^2+2(a-2)x-a^2+8\leq 4(3-a) . The function f ( x ) f(x) doesn't possess any local maximum and the function g ( x ) g(x) doesn't possess any local minimum. So A = − 4 ( a + 1 ) , B = 4 ( 3 − a ) A=-4(a+1), B=4(3-a) , and A − B = − 4 ( a + 1 + 3 − a ) = − 16 A-B=-4(a+1+3-a)=\boxed {-16}

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No, I think you interpreted this problem incorrect. Actually what H1(x) means is for any x, compute f(x) and g(x), and compare f(x),g(x), and output the bigger value.

Alice Smith - 1 year, 3 months ago

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