【寻求该题出处】已知abc是实数,函数f(x)=ax²+bx+c,g(x)=ax+b,当-1≤x≤1时,|f(x)|≤1已知abc是实数,函数f(x)=ax²+bx+c,g(x)=ax+b,当-1≤x≤1时,|f(x)|≤1 证明|c|≤12.证明,当-1≤x≤1时,|g(x)|≤2设a>0,有-
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/19 15:00:16
![【寻求该题出处】已知abc是实数,函数f(x)=ax²+bx+c,g(x)=ax+b,当-1≤x≤1时,|f(x)|≤1已知abc是实数,函数f(x)=ax²+bx+c,g(x)=ax+b,当-1≤x≤1时,|f(x)|≤1 证明|c|≤12.证明,当-1≤x≤1时,|g(x)|≤2设a>0,有-](/uploads/image/z/4485102-6-2.jpg?t=%E3%80%90%E5%AF%BB%E6%B1%82%E8%AF%A5%E9%A2%98%E5%87%BA%E5%A4%84%E3%80%91%E5%B7%B2%E7%9F%A5abc%E6%98%AF%E5%AE%9E%E6%95%B0%2C%E5%87%BD%E6%95%B0f%28x%29%3Dax%26%23178%3B%2Bbx%2Bc%2Cg%28x%29%3Dax%2Bb%2C%E5%BD%93-1%E2%89%A4x%E2%89%A41%E6%97%B6%2C%7Cf%28x%29%7C%E2%89%A41%E5%B7%B2%E7%9F%A5abc%E6%98%AF%E5%AE%9E%E6%95%B0%2C%E5%87%BD%E6%95%B0f%28x%29%3Dax%26%23178%3B%2Bbx%2Bc%2Cg%28x%29%3Dax%2Bb%2C%E5%BD%93-1%E2%89%A4x%E2%89%A41%E6%97%B6%2C%7Cf%28x%29%7C%E2%89%A41+%E8%AF%81%E6%98%8E%7Cc%7C%E2%89%A412.%E8%AF%81%E6%98%8E%2C%E5%BD%93-1%E2%89%A4x%E2%89%A41%E6%97%B6%2C%7Cg%28x%29%7C%E2%89%A42%E8%AE%BEa%26gt%3B0%2C%E6%9C%89-)
xN@_ĄhAxw$3D]`Q-[W-Ȝv+xʸĸR7.99M*'"]u6Yt̚{\FԢJUMh쳭kPrx:
&&~%鏑(vy$7X呤