【高校数学Ⅲ】1-1-1 分数関数｜問題集

\(\displaystyle =\frac{3(x-2)-1}{x-2}\)
\(\displaystyle =-\frac{1}{x-2}+3\)

\(\displaystyle =\frac{-2(x-1)+3}{x-1}\)
\(\displaystyle =\frac{3}{x-1}-2\)

\(\displaystyle =\frac{2(2x+1)+1}{2x+1}\)
\(\displaystyle =\frac{1}{2x+1}+2\)

\(\displaystyle =\frac{2(x-3)+6}{x-3}\)
\(\displaystyle =\frac{6}{x-3}+2\) よって、値域は
\(-4\leqq y\leqq 0\)

\(\displaystyle =\frac{(x-3)+1}{x-3}\)
\(\displaystyle =\frac{1}{x-3}+1\) よって、値域は
\(\displaystyle y\geqq \frac{3}{2}\)

求める範囲は\(\displaystyle y=\frac{1}{x}-2\)のグラフが直線\(y=-2x+1\)より上にある\(x\)の値の範囲なので、
\(\displaystyle 0< x\leqq \frac{1}{2},1\leqq x\)

求める範囲は\(\displaystyle y=\frac{2}{x+2}\)のグラフが直線\(y=x+3\)より下にある\(x\)の値の範囲なので、
\(\displaystyle -4\leqq x< -2,-1\leqq x\)