Given the function f(x) = 20(1.25)^x . what is the average rate of change between f(2) and f(5)

Thus, f(b)−f(a)b−a=20(54)(6103100)−(20(54)(1254))6103100−(1254)=−58207660913467407226562517167001203595951472642–√5–√4+542101086242752217003726400434970855712890625197922048572373973475376871275743307366424750⋅53100.

Answer: the average rate of change is −58207660913467407226562517167001203595951472642–√5–√4+542101086242752217003726400434970855712890625197922048572373973475376871275743307366424750⋅53100≈550754.870532511

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jdoe0001 jdoe0001 · Answer 2 · 2020-01-07T21:06:37+01:00

[tex]\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)= 20(1.25)^x\implies f(x) = 20\left( \cfrac{5}{4} \right)^x\qquad \begin{cases} x_1=2\\ x_2=5 \end{cases}\implies \cfrac{f(5)-f(2)}{5-2}[/tex]

[tex]\bf \cfrac{20\left( \frac{5}{4} \right)^5~~-~~20\left( \frac{5}{4} \right)^2}{3}\implies \cfrac{20\cdot \frac{3125}{1024}~~-~~20\cdot \frac{25}{16}}{3}\implies \cfrac{\frac{5\cdot 3125}{256}-\frac{5\cdot 25}{4}}{3} \\\\\\ \cfrac{\frac{15625}{256}-\frac{125}{4}}{3}\implies \cfrac{\frac{7625}{256}}{3}\implies \cfrac{7625}{256}\cdot \cfrac{1}{3}\implies \cfrac{7625}{768}\implies 9.92838541\overline{6}[/tex]