已知$f(x)=ax^2+bx-dfrac{1}{4}$,若存在$a,bin R$,使得对于任意的$xin[0,7],|f(x)|le2$恒成立,求$|a|$的最大值____
提示:
$|ax^2+bx-dfrac{1}{4}|le2,$得$-dfrac{7}{4x}le ax+ble dfrac{9}{4x}$结合图像,
$y=ax+b$的函数图像介于$y=-dfrac{7}{4x} extbf{与}y=dfrac{9}{4x}$的图像之间,要求$|a|$的最大值.
显然只需考虑$y=ax+b$过$(7,-dfrac{1}{4})$ 时的斜率,联立方程组
egin{equation*}
left{ egin{aligned}
y &= a(x-7)-dfrac{1}{4} \
y&=dfrac{9}{4x}
end{aligned}
ight.
end{equation*}
令$Delta=(28a+1)^2+144a=0$,得$a=-dfrac{1}{4}veedfrac{-1}{196} $所以$|a|$的最大值为$dfrac{1}{4}$