【一轮2-2】求函数的解析式

【一轮2-2】求函数的解析式

年级:高一,高二,高三
年份:2020
类型:一轮复习
题型:填空题,解答题
难度:★★★
时长:07:36
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本课程讲解求解函数解析式的4种方法,分别是:待定系数法换元法配凑法函数方程法

例1(待定系数法). f\left( x \right)为二次函数且f\left( 0 \right) =3f\left( x+2 \right) -f\left( x \right) =4x+2,求f\left( x \right)。【f\left( x \right) =x^2-x+3

解:设f\left( x \right) =ax^2+bx+c\left( a\ne 0 \right),则f\left( 0 \right) =c=3

f\left( x+2 \right) -f\left( x \right) =a\left( x+2 \right) ^2+b\left( x+2 \right) -ax^2-bx=4ax+4a+2b=4x+2

\begin{cases} 4a=4\\    4a+2b=2\\\end{cases},解得\begin{cases} a=1\\      b=-1\\\end{cases},∴f\left( x \right) =x^2-x+3

例2(换元法). 已知f\left( \ddfrac{2}{x}+1 \right) =\lg x,求f\left( x \right)。【f\left( x \right) =\lg \dfrac{2}{x-1}

解:令t=\dfrac{2}{x}+1,则t\ne 1x=\dfrac{2}{t-1},∴f\left( t \right) =\lg \dfrac{2}{t-1}\,\,\left( t\ne 1 \right)

f\left( x \right) =\lg \dfrac{2}{x-1}\,\,\left( x\ne 1 \right)

例3(配凑法).已知f\left( x+\dfrac{1}{x} \right) =x^2+\dfrac{1}{x^2},则f\left( x \right) =      。【x^2-2 \left( x\leqslant -2 \text{或} x\geqslant 2 \right)

解:∵f\left( x+\dfrac{1}{x} \right) =x^2+\dfrac{1}{x^2}=\left( x+\dfrac{1}{x} \right) ^2-2

f\left( x \right) =x^2-2 \left( x\leqslant -2 \text{或} x\geqslant 2 \right)

例4(函数方程法). 已知f\left( x \right)满足2f\left( x \right) +f\left( \dfrac{1}{x} \right) =3x-1,则f\left( x \right) =        。【2x-\dfrac{1}{x}-\dfrac{1}{3}\left( x\ne 0 \right)

解:\begin{cases}     2f\left( x \right) +f\left( \dfrac{1}{x} \right) =3x-1\cdots \cdots\\    2f\left( \dfrac{1}{x} \right) +f\left( x \right) =\dfrac{3}{x}-1\cdots \cdots\\\end{cases},①\times 2-②得:3f\left( x \right) =6x-\dfrac{3}{x}-1

f\left( x \right) =2x-\dfrac{1}{x}-\dfrac{1}{3}\,\,\left( x\ne 0 \right)

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