問題の解答
検索用コード（LaTeX）
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% 例題I3.1.1：関数の値$f(a)$（One More）★
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関数$f(x)=3x^2-5x+2$について，次の値を求めよ．
(1) $f(1)$
(2) $f(0)$
(3) $f(\frac{4}{3})$
(4) $f(2a)$
(5) $f(3a-2)$
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% 例題I3.1.1の解答（One More）★
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(1) $f(x)$に$x=1$を代入すると，
\[
f(1)=3\cdot 1^2-5\cdot 1+2=3-5+2=0
\]
(2) $f(x)$に$x=0$を代入すると，
\[
f(0)=3\cdot 0^2-5\cdot 0+2=0-0+2=2
\]
(3) $f(x)$に$x=\frac{4}{3}$を代入すると，
\[
f(\frac{4}{3})=3\cdot(\frac{4}{3})^2-5\cdot\frac{4}{3}+2=\frac{48}{9}-\frac{20}{3}+2=\frac{16}{3}-\frac{20}{3}+2=\frac{2}{3}
\]
(4) $f(x)$に$x=2a$を代入すると，
\[
f(2a)=3\cdot(2a)^2-5\cdot(2a)+2=12a^2-10a+2
\]
(5) $f(x)$に$x=3a-2$を代入すると，
\begin{align*}
f(3a-2)&=3(3a-2)^2-5(3a-2)+2\\
&=3(9a^2-12a+4)-15a+10+2\\
&=27a^2-36a+12-15a+12\\
&=27a^2-51a+24
\end{align*}
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% 問題I3.1.1：関数の値$f(a)$（One More）★
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関数$f(x)=4x^2-x+5$について，次の値を求めよ．
(1) $f(2)$
(2) $f(-\frac{3}{2})$
(3) $f(3a)$
(4) $f(2a+1)$
(5) $f(a^2-1)$
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% 問題I3.1.1の解答（One More）★
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(1) $f(x)$に$x=2$を代入すると，
\[
f(2)=4\cdot 2^2-2+5=16-2+5=19
\]
(2) $f(x)$に$x=-\frac{3}{2}$を代入すると，
\[
f(-\frac{3}{2})=4\cdot(-\frac{3}{2})^2-(-\frac{3}{2})+5=4\cdot\frac{9}{4}+\frac{3}{2}+5=\frac{31}{2}
\]
(3) $f(x)$に$x=3a$を代入すると，
\[
f(3a)=4\cdot(3a)^2-(3a)+5=36a^2-3a+5
\]
(4) $f(x)$に$x=2a+1$を代入すると，
\begin{align*}
f(2a+1)&=4(2a+1)^2-(2a+1)+5\\
&=4(4a^2+4a+1)-(2a+1)+5\\
&=16a^2+16a+4-2a-1+5\\
&=16a^2+14a+8
\end{align*}
(5) $f(x)$に$x=a^2-1$を代入すると，
\begin{align*}
f(a^2-1)&=4(a^2-1)^2-(a^2-1)+5\\
&=4(a^4-2a^2+1)-a^2+1+5\\
&=4a^4-8a^2+4-a^2+1+5\\
&=4a^4-9a^2+10
\end{align*}
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