Problems from The average change

Development:

Using the definition $$AC={\displaystyle \frac{y}{x}}={\displaystyle \frac{f(a+x)-f(a)}{x}}$$ Given the interval $[0,10]$,

$\mathrm{\Delta }y=f(10)-f(0)=({10}^2+10)-0=110$

$\mathrm{\Delta }x=10-0=10$

$AC={\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}={\displaystyle \frac{110}{10}}=11$

Given the interval $[0,2]$

$\mathrm{\Delta }y=f(2)-f(0)=(2^2+2)-0=6$

$\mathrm{\Delta }x=2-0=2$

$AC={\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}={\displaystyle \frac{6}{2}}=3$

Solution:

Interval $[0,10]:AC=11$

Interval $[0,2]:AC=3$

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