2次関数 $f(x) = 2x^2 - 3x + 4$ について、以下の値を求めます。 (1) $f(2)$ (2) $f(0)$ (3) $f(-1)$ (4) $f(\frac{1}{2})$ (5) $f(-\sqrt{2})$ (6) $f(1 - \sqrt{3})$ (7) $f(-a)$ (8) $f(a+2)$ (9) $f(a^2)$

1. 問題の内容

2次関数 $f(x) = 2x^2 - 3x + 4$ について、以下の値を求めます。

(1) $f(2)$

(2) $f(0)$

(3) $f(-1)$

(4) $f(\frac{1}{2})$

(5) $f(-\sqrt{2})$

(6) $f(1 - \sqrt{3})$

(7) $f(-a)$

(8) $f(a+2)$

(9) $f(a^2)$

2. 解き方の手順

与えられた関数 $f(x) = 2x^2 - 3x + 4$ に、それぞれの $x$ の値を代入して計算します。

(1) $f(2) = 2(2)^2 - 3(2) + 4 = 2(4) - 6 + 4 = 8 - 6 + 4 = 6$

(2) $f(0) = 2(0)^2 - 3(0) + 4 = 0 - 0 + 4 = 4$

(3) $f(-1) = 2(-1)^2 - 3(-1) + 4 = 2(1) + 3 + 4 = 2 + 3 + 4 = 9$

(4) $f(\frac{1}{2}) = 2(\frac{1}{2})^2 - 3(\frac{1}{2}) + 4 = 2(\frac{1}{4}) - \frac{3}{2} + 4 = \frac{1}{2} - \frac{3}{2} + \frac{8}{2} = \frac{1 - 3 + 8}{2} = \frac{6}{2} = 3$

(5) $f(-\sqrt{2}) = 2(-\sqrt{2})^2 - 3(-\sqrt{2}) + 4 = 2(2) + 3\sqrt{2} + 4 = 4 + 3\sqrt{2} + 4 = 8 + 3\sqrt{2}$

(6) $f(1 - \sqrt{3}) = 2(1 - \sqrt{3})^2 - 3(1 - \sqrt{3}) + 4 = 2(1 - 2\sqrt{3} + 3) - 3 + 3\sqrt{3} + 4 = 2(4 - 2\sqrt{3}) - 3 + 3\sqrt{3} + 4 = 8 - 4\sqrt{3} - 3 + 3\sqrt{3} + 4 = 9 - \sqrt{3}$

(7) $f(-a) = 2(-a)^2 - 3(-a) + 4 = 2a^2 + 3a + 4$

(8) $f(a+2) = 2(a+2)^2 - 3(a+2) + 4 = 2(a^2 + 4a + 4) - 3a - 6 + 4 = 2a^2 + 8a + 8 - 3a - 6 + 4 = 2a^2 + 5a + 6$

(9) $f(a^2) = 2(a^2)^2 - 3(a^2) + 4 = 2a^4 - 3a^2 + 4$

3. 最終的な答え

(1) $f(2) = 6$

(2) $f(0) = 4$

(3) $f(-1) = 9$

(4) $f(\frac{1}{2}) = 3$

(5) $f(-\sqrt{2}) = 8 + 3\sqrt{2}$

(6) $f(1 - \sqrt{3}) = 9 - \sqrt{3}$

(7) $f(-a) = 2a^2 + 3a + 4$

(8) $f(a+2) = 2a^2 + 5a + 6$

(9) $f(a^2) = 2a^4 - 3a^2 + 4$