Należy obliczyć iloraz różnicowy:

$\frac{\Delta y}{\Delta x}$

funkcji f w punkcie x₀.

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a)

$f\left(x\right)=\frac{1}{x},x_0=3,\Delta x=2$

Wiemy, że:

$$\begin{gathered} \frac{\Delta y}{\Delta x}=\frac{f\left(x_0+\Delta x\right)-f\left(x_0\right)}{\Delta x}= \\ =\frac{f\left(3+2\right)-f\left(3\right)}{2}= \\ =\frac{\frac{1}{5}-\frac{1}{3}}{2}=\frac{\frac{3}{15}-\frac{5}{15}}{2}=\frac{-\frac{2}{15}}{2}= \\ =-\frac{2}{15}\cdot \frac{1}{2}=-\frac{1}{15} \end{gathered}$$

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b)

$f\left(x\right)=\frac{1}{x^2},x_0=1,\Delta x=2$

$$\begin{gathered} \frac{\Delta y}{\Delta x}=\frac{f\left(x_0+\Delta x\right)-f\left(x_0\right)}{\Delta x}= \\ =\frac{f\left(1+2\right)-f\left(1\right)}{2}= \\ =\frac{\frac{1}{3^2}-\frac{1}{1^2}}{2}=\frac{\frac{1}{9}-\frac{1}{1}}{2}=\frac{-\frac{8}{9}}{2}= \\ =-\frac{8}{9}\cdot \frac{1}{2}=-\frac{4}{9} \end{gathered}$$

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c)

$f\left(x\right)=x^2-4,x_0=2,\Delta x=-1$

$$\begin{gathered} \frac{\Delta y}{\Delta x}=\frac{f\left(x_0+\Delta x\right)-f\left(x_0\right)}{\Delta x}= \\ =\frac{f\left(2-1\right)-f\left(2\right)}{-1}= \\ =\frac{1^2-4-\left(2^2-4\right)}{-1}= \\ =\frac{-3-0}{-1}=\frac{-3}{-1}=3 \end{gathered}$$

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d)

$f\left(x\right)=3x^2+2x-1,x_0=0,\Delta x=-1$

$$\begin{gathered} \frac{\Delta y}{\Delta x}=\frac{f\left(x_0+\Delta x\right)-f\left(x_0\right)}{\Delta x}= \\ =\frac{f\left(0-1\right)-f\left(0\right)}{-1}= \\ =\frac{3\cdot {\left(-1\right)}^2+2\cdot \left(-1\right)-1+1}{-1}= \\ =\frac{3-2}{-1}=\frac{1}{-1}=-1 \end{gathered}$$

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e)

$f\left(x\right)=\frac{1}{x^2+1},x_0=0,\Delta x=1$

$$\begin{gathered} \frac{\Delta y}{\Delta x}=\frac{f\left(x_0+\Delta x\right)-f\left(x_0\right)}{\Delta x}= \\ =\frac{f\left(0+1\right)-f\left(0\right)}{1}= \\ =\frac{\frac{1}{1^2+1}-\frac{1}{0^2+1}}{1}=\frac{\frac{1}{2}-1}{1}=-\frac{1}{2} \end{gathered}$$

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f)

$f\left(x\right)=\frac{x^2-4}{x^2+4},x_0=2,\Delta x=-1$

$$\begin{gathered} \frac{\Delta y}{\Delta x}=\frac{f\left(x_0+\Delta x\right)-f\left(x_0\right)}{\Delta x}= \\ =\frac{f\left(2-1\right)-f\left(2\right)}{-1}= \\ =\frac{\frac{1^2-4}{1^2+4}-\frac{2^2-4}{2^2+4}}{-1}=\frac{\frac{-3}{5}-0}{-1}=\frac{3}{5} \end{gathered}$$