《2025年芝麻开花美在课堂高一数学必修第一册北师大版》

[例3] 求下列函数的解析式:
(1)(换元法或配凑法)已知$f(\sqrt{x}+1)= x - 2\sqrt{x}$,求$f(x)$;
(2)(待定系数法)已知函数$f(x)$是一次函数,若$f(f(x)) = 4x + 8$,求$f(x)$;
(3)(解方程组法)已知函数$f(x)对于任意的x都有f(x)-2f(-x)= 1 + 2x$,求$f(x)$.

答案： 解:
(1)法一:换元法 令$t=\sqrt {x}+1$,则$t≥1,x=(t-1)^{2}$.代入原式有$f(t)=(t-1)^{2}-2(t-1)=t^{2}-4t+3$,所以$f(x)=x^{2}-4x+3(x≥1).$
法二:配凑法 由已知得$f(\sqrt {x}+1)=x+2\sqrt {x}+1-4\sqrt {x}-4+3=(\sqrt {x}+1)^{2}-4(\sqrt {x}+1)+3$.
因为$\sqrt {x}+1≥1$,所以$f(x)=x^{2}-4x+3(x≥1).$
(2)设$f(x)=ax+b(a≠0)$,
则$f(f(x))=f(ax+b)=a(ax+b)+b=a^{2}x+ab+b$.
因为$f(f(x))=4x+8$,所以$a^{2}x+ab+b=4x+8$,
即$\left\{\begin{array}{l} a^{2}=4,\\ ab+b=8,\end{array}\right. $解得$\left\{\begin{array}{l} a=2,\\ b=\frac {8}{3}\end{array}\right. $或$\left\{\begin{array}{l} a=-2,\\ b=-8.\end{array}\right. $
所以$f(x)=2x+\frac {8}{3}$或$f(x)=-2x-8.$
(3)由题意,在$f(x)-2f(-x)=1+2x$中,以-x代替x可得$f(-x)-2f(x)=1-2x.$
联立$\left\{\begin{array}{l} f(x)-2f(-x)=1+2x,\\ f(-x)-2f(x)=1-2x,\end{array}\right. $解得$f(x)=\frac {2}{3}x-1.$

4.根据下列条件,求$f(x)$的解析式.
(1)已知$f(x)满足f(x + 1)= x^2+4x + 1$;
(2)已知$f(x)$是一次函数,且满足$3f(x + 1)-f(x)= 2x + 9$;
(3)已知$f(x)满足2f(\frac{1}{x})+f(x)= x(x\neq0)$.

答案： 解:
(1)令$t=x+1$,则$x=t-1$.
故$f(t)=(t-1)^{2}+4(t-1)+1=t^{2}+2t-2$.
所以$f(x)=x^{2}+2x-2.$
(2)设$f(x)=kx+b(k≠0)$,因为$3f(x+1)-f(x)=2x+9$,所以$3k(x+1)+3b-kx-b=2x+9$.即$2kx+3k+2b=2x+9$.所以$\left\{\begin{array}{l} 2k=2,\\ 3k+2b=9,\end{array}\right. $解得$\left\{\begin{array}{l} k=1,\\ b=3.\end{array}\right. $所以$f(x)=x+3.$
(3)因为$2f(\frac {1}{x})+f(x)=x(x≠0)$①,所以$2f(x)+f(\frac {1}{x})=\frac {1}{x}$②.$2×②-①$,得$3f(x)=\frac {2}{x}-x$,所以$f(x)=\frac {2}{3x}-\frac {x}{3}(x≠0).$