【高校数学Ⅲ】1-1-1 分数関数｜問題集
1.次の関数のグラフを描きなさい。
(1)\(\displaystyle y=\frac{2}{x+1}-3\)
(2)\(\displaystyle y=\frac{3x-7}{x-2}\)
\(\displaystyle =\frac{3(x-2)-1}{x-2}\)
\(\displaystyle =-\frac{1}{x-2}+3\)
\(\displaystyle =-\frac{1}{x-2}+3\)
(3)\(\displaystyle y=\frac{-2x+5}{x-1}\)
\(\displaystyle =\frac{-2(x-1)+3}{x-1}\)
\(\displaystyle =\frac{3}{x-1}-2\)
\(\displaystyle =\frac{3}{x-1}-2\)
(4)\(\displaystyle y=\frac{4x+3}{2x+1}\)
\(\displaystyle =\frac{2(2x+1)+1}{2x+1}\)
\(\displaystyle =\frac{1}{2x+1}+2\)
\(\displaystyle =\frac{1}{2x+1}+2\)
2.次の関数の値域を求めなさい。
(1)\(\displaystyle y=\frac{2x}{x-3}(0\leqq x\leqq 2)\)
\(\displaystyle =\frac{2(x-3)+6}{x-3}\)
\(\displaystyle =\frac{6}{x-3}+2\)
よって、値域は
\(-4\leqq y\leqq 0\)
\(\displaystyle =\frac{6}{x-3}+2\)
よって、値域は
\(-4\leqq y\leqq 0\)
(2)\(\displaystyle y=\frac{x-2}{x-3}(3< x\leqq 5)\)
\(\displaystyle =\frac{(x-3)+1}{x-3}\)
\(\displaystyle =\frac{1}{x-3}+1\)
よって、値域は
\(\displaystyle y\geqq \frac{3}{2}\)
\(\displaystyle =\frac{1}{x-3}+1\)
よって、値域は
\(\displaystyle y\geqq \frac{3}{2}\)
3.次の不等式を解きなさい。
(1)\(\displaystyle \frac{1}{x}-2\geqq -2x+1\)
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、
\(\displaystyle 0< x\leqq \frac{1}{2},1\leqq x\)
(2)\(\displaystyle \frac{2}{x+2}\leqq x+3\)
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、
\(\displaystyle -4\leqq x< -2,-1\leqq x\)
次の学習に進もう！