【高校数学Ⅱ】6-1-1 微分係数｜問題集

\(\displaystyle \frac{f(3)-f(0)}{3-0}\)
\(\displaystyle =\frac{-8-1}{3}\)
\(=-3\)

\(\displaystyle \frac{f(b)-f(a)}{b-a}\)
\(\displaystyle =\frac{2(b-a)}{b-a}\)
\(=2\)

\(\displaystyle \frac{f(2+h)-f(2)}{(2+h)-2}\)
\(\displaystyle =\frac{-h^2-4h}{h}\)
\(=-h-4\)

\(\displaystyle \frac{f(4)-f(1)}{4-1}\)
\(\displaystyle =\frac{0-(-3)}{3}\)
\(=1\)

\(\displaystyle \frac{f(1)-f(-2)}{1-(-2)}\)
\(\displaystyle =\frac{-1-2}{3}\)
\(=-1\)

\(=3\)

\(=-2\)

\(\displaystyle f'(0)=\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\lim_{h \to 0}\frac{2h-3-(-3)}{h}\)
\(\ \ \ \ \ \ \ \ =2\)

\(\displaystyle f'(1)=\lim_{h \to 0}\frac{f(1+h)-f(1)}{h}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\lim_{h \to 0}\frac{h^2+2h}{h}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\lim_{h \to 0}(h+2)\)
\(\ \ \ \ \ \ \ \ =2\)

\(\displaystyle f'(3)=\lim_{h \to 0}\frac{f(3+h)-f(3)}{h}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\lim_{h \to 0}\frac{h^2+2h}{h}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\lim_{h \to 0}(h+2)\)
\(\ \ \ \ \ \ \ \ =2\)

\(\displaystyle f'(-1)=\lim_{h \to 0}\frac{f(-1+h)-f(-1)}{h}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\lim_{h \to 0}\frac{h^2-2h}{h}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\lim_{h \to 0}(h-2)\)
\(\ \ \ \ \ \ \ \ =-2\)