1.次の関数の増減を調べなさい。
(1)\(f(x)=x-e^x\)
\(\displaystyle f'(x)=1-e^x\)
増減表にまとめると、
| \(x\) | \(\cdots\) | \(0\) | \(\cdots\) |
| \(f'(x)\) | \(+\) | \(0\) | \(-\) |
| \(f(x)\) | \(\nearrow\) | \(0\) | \(\searrow\) |
よって、\(f(x)\)は
\(\displaystyle x<0\)で増加し、\(\displaystyle x\geqq0\)で減少する。
(2)\(f(x)=x-\log x\)
関数\(f(x)\)の定義域は\(x>0\)である。
\(\displaystyle f'(x)=1-\frac{1}{x}\)
\(\displaystyle =\frac{x-1}{x}\)
増減表にまとめると、
| \(x\) | \(0\) | \(\cdots\) | \(1\) | \(\cdots\) |
| \(f'(x)\) | \(-\) | \(0\) | \(+\) | |
| \(f(x)\) | \(0\) | \(\searrow\) | \(1\) | \(\nearrow\) |
よって、\(f(x)\)は
\(0< x\leqq1\)で減少し、\(x\geqq1\)で増加する。
2.次の関数の極値を求めなさい。
(1)\(f(x)=x^2e^{-x}\)
\(f'(x)=2xe^{-x}-x^2e^{-x}\)
\(=(2-x)xe^{-x}\)
増減表にまとめると、
| \(x\) | \(\cdots\) | \(0\) | \(\cdots\) | \(2\) | \(\cdots\) |
| \(f'(x)\) | \(-\) | \(0\) | \(+\) | \(0\) | \(-\) |
| \(f(x)\) | \(\searrow\) | \(0\) | \(\nearrow\) | \(\displaystyle \frac{4}{e^2}\) | \(\searrow\) |
したがって、
\(x=0\)のとき、極小値\(0\)
\(x=2\)のとき、極大値\(\displaystyle \frac{4}{e^2}\)
(2)\(f(x)=x\log x\)
関数\(f(x)\)の定義域は\(x>0\)である。
\(\displaystyle f'(x)=\log x+x・\frac{1}{x}\)
\(=\log x+1\)
増減表にまとめると、
| \(x\) | \(0\) | \(\cdots\) | \(\displaystyle \frac{1}{e}\) | \(\cdots\) |
| \(f'(x)\) | \(-\) | \(0\) | \(+\) | |
| \(f(x)\) | \(\searrow\) | \(\displaystyle -\frac{1}{e}\) | \(\nearrow\) |
したがって、
\(\displaystyle x=\frac{1}{e}\)のとき、極小値\(\displaystyle -\frac{1}{e}\)
(3)\(\displaystyle f(x)=x+\frac{2}{x}\)
関数\(f(x)\)の定義域は\(x\neq0\)である。
\(\displaystyle f'(x)=1-\frac{2}{x^2}\)
増減表にまとめると、
| \(x\) | \(\cdots\) | \(-\sqrt{2}\) | \(\cdots\) | \(0\) | \(\cdots\) | \(\sqrt{2}\) | \(\cdots\) |
| \(f'(x)\) | \(+\) | \(0\) | \(-\) | \(-\) | \(0\) | \(+\) | |
| \(f(x)\) | \(\nearrow\) | \(-2\sqrt{2}\) | \(\searrow\) | \(\searrow\) | \(2\sqrt{2}\) | \(\nearrow\) |
したがって、
\(x=-\sqrt{2}\)のとき、極大値\(-2\sqrt{2}\)
\(x=\sqrt{2}\)のとき、極小値\(2\sqrt{2}\)
(4)\(\displaystyle f(x)=|x|(x+1)\)
(1)\(x>0\)のとき、
\(f(x)=x(x+1)\)
\(f'(x)=2x+1\)
(2)\(x<0\)のとき、
\(f(x)=-x(x+1)\)
\(f'(x)=-2x-1\)
増減表にまとめると、
| \(x\) | \(\cdots\) | \(\displaystyle -\frac{1}{2}\) | \(\cdots\) | \(0\) | \(\cdots\) |
| \(f'(x)\) | \(+\) | \(0\) | \(-\) | \(+\) | |
| \(f(x)\) | \(\nearrow\) | \(\displaystyle \frac{1}{4}\) | \(\searrow\) | \(1\) | \(\nearrow\) |
したがって、
\(\displaystyle x=-\frac{1}{2}\)のとき、極大値\(\displaystyle \frac{1}{4}\)
\(x=0\)のとき、極小値\(1\)
(5)\(\displaystyle f(x)=|x|\sqrt{x+2}\)
関数\(f(x)\)の定義域は\(x\leqq-2\)である。
(1)\(x>0\)のとき、
\(f(x)=x\sqrt{x+2}\)
\(\displaystyle f'(x)=\sqrt{x+2}+\frac{x}{2\sqrt{x+2}}\)
\(\displaystyle =\frac{3x+4}{2\sqrt{x+2}}\)
(2)\(x<0\)のとき、
\(f(x)=-x\sqrt{x+2}\)
\(\displaystyle f'(x)=-\sqrt{x+2}-\frac{x}{2\sqrt{x+2}}\)
\(\displaystyle =\frac{-3x-4}{2\sqrt{x+2}}\)
増減表にまとめると、
| \(x\) | \(-2\) | \(\cdots\) | \(\displaystyle -\frac{4}{3}\) | \(\cdots\) | \(0\) | \(\cdots\) |
| \(f'(x)\) | \(+\) | \(0\) | \(-\) | \(+\) | ||
| \(f(x)\) | \(0\) | \(\nearrow\) | \(\displaystyle \frac{4\sqrt{6}}{9}\) | \(\searrow\) | \(0\) | \(\nearrow\) |
したがって、
\(\displaystyle x=-\frac{4}{3}\)のとき、極大値\(\displaystyle \frac{4\sqrt{6}}{9}\)
\(x=0\)のとき、極小値\(0\)
3.次の関数と極値をとるとき、定数\(a\)を求めなさい。また、\(f(x)\)の極値を求めなさい。
(1)\(\displaystyle f(x)=x+\frac{a}{x}\)
極値:\(x=1\)
関数\(f(x)\)の定義域は\(x\leqq-2\)である。
\(\displaystyle f'(x)=1-\frac{a}{x^2}\)
\(\displaystyle f'(1)=1-\frac{a}{1^2}=0\)
よって、\(a=1\)
\(\displaystyle f(x)=x+\frac{1}{x}\)
\(\displaystyle f'(x)=1-\frac{1}{x^2}\)
増減表にまとめると、
| \(x\) | \(\cdots\) | \(-1\) | \(\cdots\) | \(0\) | \(\cdots\) | \(1\) | \(\cdots\) |
| \(f'(x)\) | \(+\) | \(0\) | \(-\) | \(-\) | \(0\) | \(+\) | |
| \(f(x)\) | \(\nearrow\) | \(-2\) | \(\searrow\) | \(\searrow\) | \(2\) | \(\nearrow\) |
したがって、
\(x=-1\)のとき、極大値\(-2\)
\(x=1\)のとき、極小値\(2\)
(2)\(\displaystyle f(x)=\frac{2x-a}{x^2+2}\)
極値:\(x=-1\)
\(\displaystyle f'(x)=\frac{2(x^2+2)-2x(2x-a)}{(x^2+2)^2}\)
\(\displaystyle =\frac{-2x^2+2ax+4}{(x^2+2)^2}\)
\(\displaystyle f'(-1)=\frac{-2-2a+4}{(1^2+2)^2}=0\)
よって、\(a=1\)
\(\displaystyle f(x)=\frac{2x-1}{x^2+2}\)
\(\displaystyle f'(x)=\frac{-2(x+1)(x-2)}{(x^2+2)^2}\)
増減表にまとめると、
| \(x\) | \(\cdots\) | \(-1\) | \(\cdots\) | \(2\) | \(\cdots\) |
| \(f'(x)\) | \(-\) | \(0\) | \(+\) | \(0\) | \(-\) |
| \(f(x)\) | \(\searrow\) | \(-1\) | \(\nearrow\) | \(\displaystyle \frac{1}{2}\) | \(\searrow\) |
したがって、
\(x=-1\)のとき、極小値\(-1\)
\(x=2\)のとき、極大値\(\displaystyle \frac{1}{2}\)
(3)\(\displaystyle f(x)=\frac{x}{x^2+a}\)
極値:\(x=-2\)
\(\displaystyle f'(x)=\frac{(x^2+a)-x・2x}{(x^2+a)^2}\)
\(\displaystyle =\frac{-x^2+a}{(x^2+a)^2}\)
\(\displaystyle f'(-2)=\frac{-(-2)^2+a}{((-2)^2+a)^2}=0\)
よって、\(a=4\)
\(\displaystyle f(x)=\frac{x}{x^2+4}\)
\(\displaystyle f'(x)=\frac{-(x+2)(x-2)}{(x^2+4)^2}\)
増減表にまとめると、
| \(x\) | \(\cdots\) | \(-2\) | \(\cdots\) | \(-2\) | \(\cdots\) |
| \(f'(x)\) | \(-\) | \(0\) | \(+\) | \(0\) | \(-\) |
| \(f(x)\) | \(\searrow\) | \(\displaystyle -\frac{1}{4}\) | \(\nearrow\) | \(\displaystyle \frac{1}{4}\) | \(\searrow\) |
したがって、
\(x=-2\)のとき、極小値\(\displaystyle -\frac{1}{4}\)
\(x=2\)のとき、極大値\(\displaystyle \frac{1}{4}\)
4.次の関数の最大値と最小値を求めなさい。
(1)\(f(x)=(1+\cos x)\sin x\)
\((0\leqq x\leqq2\pi)\)
\(f'(x)=(-\sin x)\sin x+(1+\cos x)\cos x\)
\(=-\sin^2x+\cos x+\cos^2x\)
\(=2\cos^2x+\cos x-1\)
\(=(2\cos x-1)(\cos x+1))\)
増減表にまとめると、
| \(x\) | \(0\) | \(\cdots\) | \(\displaystyle \frac{\pi}{3}\) | \(\cdots\) | \(\pi\) | \(\cdots\) | \(\displaystyle \frac{5\pi}{3}\) | \(\cdots\) | \(2\pi\) |
| \(f'(x)\) | \(+\) | \(0\) | \(-\) | \(0\) | \(-\) | \(0\) | \(+\) | ||
| \(f(x)\) | \(0\) | \(\nearrow\) | \(\displaystyle \frac{3\sqrt{3}}{4}\) | \(\searrow\) | \(0\) | \(\searrow\) | \(\displaystyle -\frac{3\sqrt{3}}{4}\) | \(\nearrow\) | \(0\) |
したがって、
\(\displaystyle x=\frac{\pi}{3}\)のとき、最大値\(\displaystyle \frac{3\sqrt{3}}{4}\)
\(\displaystyle x=\frac{5\pi}{3}\)のとき、最小値\(\displaystyle -\frac{3\sqrt{3}}{4}\)
(2)\(\displaystyle f(x)=\frac{4-3x}{x^2+1}\)
\((1\leqq x\leqq4)\)
\(\displaystyle f'(x)=\frac{-3(x^2+1)-2x(4-3x)}{(x^2+1)^2}\)
\(\displaystyle =\frac{3x^2-8x-3}{(x^2+1)^2}\)
\(\displaystyle =\frac{(3x+1)(x-3)}{(x^2+1)^2}\)
増減表にまとめると、
| \(x\) | \(1\) | \(\cdots\) | \(3\) | \(\cdots\) | \(4\) |
| \(f'(x)\) | \(-\) | \(0\) | \(+\) | ||
| \(f(x)\) | \(\displaystyle \frac{1}{2}\) | \(\searrow\) | \(\displaystyle -\frac{1}{2}\) | \(\nearrow\) | \(\displaystyle -\frac{8}{17}\) |
したがって、
\(x=1\)のとき、最大値\(\displaystyle \frac{1}{2}\)
\(x=3\)のとき、最小値\(\displaystyle -\frac{1}{2}\)