14　関数のグラフ（増減・凹凸）

\(x\)

\(y\)

\(y = e^x\)

\(y = \log x\)

\(x\)

\(y\)

\(x\)

\(y\)

\(x = \)

\(f'(x) = \)

\(f''(x) = \)

1. \(f(x) = x^4 - 2x^2 - 1\)　解答 隠す

\(f'(x) \begin{array}[t]{l} = 4x^3 - 4x \\ = 4x(x^2 - 1) \\ = 4x(x + 1)(x -1) \end{array}\)

\(f''(x) \begin{array}[t]{l} = 12x^2 - 4 \\ = 4(\sqrt{3}\,x + 1)(\sqrt{3}\,x - 1) \end{array}\)

\(\begin{array}[t]{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline x & \cdots & -1 & \cdots & -\frac{1}{\sqrt{3}} & \cdots & 0 & \cdots & \frac{1}{\sqrt{3}} & \cdots & 1 & \cdots \\ \hline f'(x) & - & 0 & + & + & + & 0 & - & - & - & 0 & + \\ \hline f''(x) & + & + & + & 0 & - & - & - & 0 & + & + & + \\ \hline f(x) & \searrow_{\cup} & -2 & \nearrow_{\cup} & -\frac{14}{9} & \nearrow_{\cap} & -1 & \searrow_{\cap} & -\frac{14}{9} & \searrow_{\cup} & -2 & \nearrow_{\cup} \\ \hline \end{array}\)

\(x\)

\(y\)

極大点：\(\left(0\mbox{，}-1\right)\)　極小点：\(\left(\pm 1\mbox{，}-2\right)\)

変曲点：\(\displaystyle \left(\pm \frac{1}{\sqrt{3}}\mbox{，}-\frac{14}{9}\right)\)

2. \(f(x) = \displaystyle \frac{x}{x^2 + 1}\)　解答 隠す

\(f'(x) \begin{array}[t]{l} \displaystyle = \frac{(x)' \cdot (x^2 + 1) - x \cdot (x^2 + 1)'}{(x^2 + 1)^2} \\ \displaystyle = \frac{x^2 + 1 - x \cdot 2x}{(x^2 + 1)^2} \\ \displaystyle = \frac{-x^2 + 1}{(x^2 + 1)^2} \\ \displaystyle = -\frac{(x + 1)(x - 1)}{(x^2 + 1)^2} \end{array}\)

\(f''(x) \begin{array}[t]{l} \displaystyle = \frac{(-x^2 + 1)' \cdot (x^2 + 1)^2 - (-x^2 + 1) \cdot \{(x^2 + 1)^2\}'}{(x^2 + 1)^4} \\ \displaystyle = \frac{-2x(x^2 + 1)^2 + (x^2 - 1) \cdot \{2(x^2 + 1) \cdot (x^2 + 1)'\}}{(x^2 + 1)^4} \\ \displaystyle = -\frac{-2x(x^2 + 1) + (x^2 - 1) \cdot 2 \cdot 2x}{(x^2 + 1)^3} \\ \displaystyle = \frac{-2x^3 - 2x + 4x^3 - 4x}{(x^2 + 1)^3} \\ \displaystyle = \frac{2x^3 - 6x}{(x^2 + 1)^3} \\ \displaystyle = \frac{2x(x^2 - 3)}{(x^2 + 1)^3} \end{array}\)

\(\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline x & \cdots & -\sqrt{3} & \cdots & -1 & \cdots & 0 & \cdots & 1 & \cdots & \sqrt{3} & \cdots \\ \hline f'(x) & - & - & - & 0 & + & + & + & 0 & - & - & - \\ \hline f''(x) & - & 0 & + & + & + & 0 & - & - & - & 0 & + \\ \hline f(x) & \searrow_{\cap} & -\frac{\sqrt{3}}{4} & \searrow_{\cup} & -\frac{1}{2} & \nearrow_{\cup} & 0 & \nearrow_{\cap} & \frac{1}{2} & \searrow_{\cap} & \frac{\sqrt{3}}{4} & \searrow_{\cup} \\ \hline \end{array}\)

また，\(\left\{\begin{array}[m]{l} \displaystyle \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{\frac{1}{x}}{1 + \frac{1}{x^2}} = 0 \\ \displaystyle \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{\frac{1}{x}}{1 + \frac{1}{x^2}} = 0 \end{array}\right.\)

∴　グラフは \(x \to \pm\infty\) でとも \(x\) 軸に漸近する。

\(x\)

\(y\)

極大点：\(\displaystyle \left(1\mbox{，}\frac{1}{2}\right)\)　極小点：\(\displaystyle \left(-1\mbox{，}-\frac{1}{2}\right)\)

変曲点：\(\displaystyle \left(\pm \sqrt{3}\mbox{，}\pm \frac{\sqrt{3}}{4}\right)\mbox{，}(0\mbox{，}0)\)

3. \(f(x) = x + \sin x \quad (0 \leqq x \leqq 2\pi)\)　解答 隠す

\(\begin{array}{l} f'(x) = 1 + \cos x \\ f''(x) = -\sin x \end{array}\)

\(\begin{array}{|c|c|c|c|c|c|} \hline x & 0 & \cdots & \pi & \cdots & 2\pi \\ \hline f'(x) && + & 0 & + & \\ \hline f''(x) && - & 0 & + & \\ \hline f(x) & 0 & \nearrow_{\cap} & \pi & \nearrow_{\cup} & 2\pi \\ \hline \end{array}\)

\(x\)

\(y\)

極値なし，変曲点 \((\pi\mbox{，}\pi)\)

4. \(f(x) = (\log x)^2\)　解答 隠す

\(\displaystyle f'(x) = 2(\log x) \cdot (\log x)' = \frac{2\log x}{x}\)

\(f''(x) \begin{array}[t]{l} \displaystyle = \frac{2\{(\log x)' \cdot x - \log x \cdot (x)'\}}{x^2} \\ \displaystyle = \frac{2\{\frac{1}{x} \cdot x - \log x\}}{x^2} \\ \displaystyle = \frac{2(1 - \log x)}{x^2} \end{array}\)

\(\begin{array}{|c|c|c|c|c|c|c|} \hline x & (0) & \cdots & 1 & \cdots & e & \cdots \\ \hline f'(x) && - & 0 & + & + & + \\ \hline f''(x) && + & + & + & 0 & - \\ \hline f(x) && \searrow_{\cup} & 0 & \nearrow_{\cup} & 1 & \nearrow_{\cap} \\ \hline \end{array}\)

また，\(\displaystyle \lim_{x \to +0} f(x) = \infty\) だから，グラフは \(y\) 軸を漸近線にもつ。

\(x\)

\(y\)

極小点：\(\left(1\mbox{，}0\right)\)　変曲点：\(\left(e\mbox{，}1\right)\)

5. \(f(x) = e^{-2x^2}\)　解答 隠す

\(f'(x) = e^{-2x^2} \cdot (-2x^2)' = -4x e^{-2x^2}\)

\(f''(x) \begin{array}[t]{l} = -4\left\{(x)' \cdot e^{-2x^2} + x \cdot (e^{-2x^2})'\right\} \\ = -4\left\{e^{-2x^2} + x \cdot e^{-2x^2} \cdot (-2x^2)'\right\} \\ = -4\left(e^{-2x^2} - 4x^2 e^{-2x^2}\right) \\ = 4e^{-2x^2}(2x + 1)(2x - 1) \end{array}\)

\(\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & \cdots & -\frac{1}{2} & \cdots & 0 & \cdots & \frac{1}{2} & \cdots \\ \hline f'(x) & + & + & + & 0 & - & - & - \\ \hline f''(x) & + & 0 + & - & - & - & 0 & + \\ \hline f(x) & \nearrow_{\cup} & \frac{1}{\sqrt{e}} & \nearrow_{\cap} & 1 & \searrow_{\cap} & \frac{1}{\sqrt{e}} & \searrow_{\cup} \\ \hline \end{array}\)

また，\(\displaystyle \lim_{x \to -\infty} f(x) = \lim_{x \to \infty} f(x) = 0\) だから \(x \to \pm \infty\) のとき，\(x\) 軸に漸近する。

\(x\)

\(y\)

極大点：\(\displaystyle \left(0\mbox{，}1\right)\)　変曲点：\(\displaystyle \left(\pm \frac{1}{2}\mbox{，}\frac{1}{\sqrt{e}}\right)\)