Given f(x) = 17-х^2, what is the average rate of change in f(x) over the interval [1, 5]?

1 answer:

8 0

$\bf slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\[-0.35em] \rule{34em}{0.25pt}\\\\ f(x)=17-x^2 \qquad \begin{cases} x_1=1\\ x_2=5 \end{cases}\implies \cfrac{f(5)-f(1)}{5-1} \\\\\\ \cfrac{[17-(5)^2]~~-~~[17-(1)^2]}{4}\implies \cfrac{[-8]-[16]}{4}\implies \cfrac{-24}{4}\implies -6$

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