Given the function f (x) = (A & # 178; + 4) e ^ (X-5), G (x) = (X & # 178; + ax-2a-3) e ^ (3-x) It is proved that when a ＜ - 6, there must be x1, X2 ∈ [0,5] such that f (x1) - G (x2) > 40

Given the function f (x) = (A & # 178; + 4) e ^ (X-5), G (x) = (X & # 178; + ax-2a-3) e ^ (3-x) It is proved that when a ＜ - 6, there must be x1, X2 ∈ [0,5] such that f (x1) - G (x2) > 40

Visual inspection: if f (x) is disassembled, a & # 178; + 4, when a ＜ - 6, the range is (40, + infinity); e ^ (X-5) is in [0,5], the range is (0,1], so the maximum value of F (x) is the maximum only when x = 5, which is equal to a & # 178; + 4, the range is (40, + infinity)

The maximum value of function f (x) = x + X & # 179; / 1 + 8x & # 178; + X & # 8308

f(x)=(1/x+x)/(1/x^2+8+x^2)
Let t = x + 1 / x, then: | t | > = 2
f(x)=t/(t^2+6)
T ^ 2 + 6 > = 2 √ (T ^ 2 * 6) = 2 √ 6 | t | when | t | = √ 6, take the equal sign
So we have: | f (x)|

If f (x) = x & # 8308;, let g (x) = - QF (x) + (2q-1) x & # 178; + 1, is there a real number Q (q

Existence
Let X & sup2; = t
Then the function g (x) = - QF (x) + (2q-1) x & sup2; + 1 = - QT & sup2; + (2q-1) t + 1 t belongs to zero to positive infinity
The symmetry axis is t = 2q-1 / 2q and Q