Calculate the average rate of change of a function between two points using the equation of the function Precalculus

The average rate of change of f between -1 and 2 is:

m_{avg} = \dfrac{f\left(2\right) - f\left(-1\right)}{2 - \left(-1\right)}

We have:

f\left(2\right) = 3\left(2^3\right) - 2\left(2\right) + 4

f\left(2\right) = 24 - 4 + 4

f\left(2\right) = 24

And:

f\left(-1\right) = 3\left(\left(-1\right)^3\right) - 2\left(-1\right) + 4

f\left(-1\right) = -3 +2 + 4

f\left(-1\right) = 3

Therefore:

m_{avg} = \dfrac{24 - 3}{2 - \left(-1\right)}

m_{avg} = \dfrac{21}{3}=7

The average rate of change of f between -2 and 3 is:

m_{avg} = \dfrac{f\left(3\right) - f\left(-2\right)}{3 - \left(-2\right)}

We have:

f\left(3\right) = -2\left(3^2\right) + 3\left(3\right) + 5

f\left(3\right) = -18 +9 + 5

f\left(3\right) = -4

And:

f\left(-2\right) = -2\left(-2\right)^2 + 3\left(-2\right) + 5

f\left(-2\right) = -8 + -6 + 5

f\left(-2\right) = -9

Therefore:

m_{avg} = \dfrac{-4 - \left(-9\right)}{3 - \left(-2\right)}

m_{avg} = \dfrac{5}{5}=1

The average rate of change of f between 5 and 9 is:

m_{avg} = \dfrac{f\left(9\right) - f\left(5\right)}{9 - 5}

We have:

f\left(9\right) = \left(9 - 7\right)^2 +4

f\left(9\right) = 2^2 + 4

f\left(9\right) = 8

And:

f\left(5\right) = \left(5 - 7\right)^2 + 4

f\left(5\right) =\left(-2\right)^2 + 4

f\left(5\right) = 8

Therefore:

m_{avg} = \dfrac{8 - 8}{9 - 5}

m_{avg} = \dfrac{0}{4} = 0

The average rate of change of f between -2 and 2 is:

m_{avg} = \dfrac{f\left(2\right) - f\left(-2\right)}{2 - \left(-2\right)}

We have:

f\left(2\right) = -2\left(2\right)^3 + 4\left(2\right)^2 - 3\left(2\right)

f\left(2\right) = -2\left(8\right) + 16 - 6

f\left(2\right) = -16 + 16 - 6

f\left(2\right) = -6

And:

f\left(-2\right) = -2\left(-2\right)^3 +4\left(-2\right)^2 - 3\left(-2\right)

f\left(-2\right) = -2\left(-8\right) + 4\left(4\right) + 6

f\left(-2\right) = 16 + 16 + 6

f\left(-2\right) = 38

Therefore:

m_{avg} = \dfrac{-6 - 38}{2 - \left(-2\right)}

m_{avg} = \dfrac{-44}{4}=-11

The average rate of change of f between 2 and 3 is:

m_{avg} = \dfrac{f\left(3\right) - f\left(2\right)}{3 - 2}

We have:

f\left(3\right) = \dfrac{2}{3}\left(1 - 3\right)^3

f\left(3\right) = \dfrac{2}{3} \left(-2\right)^3

f\left(3\right) = \dfrac{2}{3} \left(-8\right)

f\left(3\right) = \dfrac{-16}{3}

And:

f\left(2\right) = \dfrac{2}{3} \left(1 - 2\right)^3

f\left(2\right) = \dfrac{2}{3}\left(-1\right)^3

f\left(2\right) = -\dfrac{2}{3}

Therefore:

m_{avg} = \dfrac{-16/3 - \left(-2/3\right)}{3 - 2}

m_{avg} = \dfrac{-14/3}{1}=-\dfrac{14}{3}

The average rate of change of f between -1 and 3 is:

m_{avg} = \dfrac{f\left(3\right) - f\left(-1\right)}{3 - \left(-1\right)}

We have:

f\left(3\right) = \dfrac{\left(1 - 3\right)^3}{4\left(3 - 2\right)^2}

f\left(3\right) = \dfrac{\left(-2\right)^3}{4\left(1\right)^2}

f\left(3\right) = \dfrac{-8}{4}

f\left(3\right) = -2

And:

f\left(-1\right) = \dfrac{\left(1 - \left(-1\right)\right)^3}{4\left(-1 - 2\right)^2}

f\left(-1\right) = \dfrac{\left(1 + 1\right)^3}{4\left(-3\right)^2}

f\left(-1\right) = \dfrac{8}{36}

which simplifies to

f\left(-1\right) = \dfrac{2}{9} .

Therefore:

m_{avg} = \dfrac{-2 - \dfrac{2}{9}}{3- \left(-1\right)}

m_{avg} = \dfrac{-\dfrac{18}{9} - \dfrac{2}{9}}{3+1}

m_{avg} = \dfrac{\dfrac{-20}{9}}{4}

m_{avg} = \dfrac{-20}{9}\cdot\dfrac{1}{4}

m_{avg} = -\dfrac{5}{9}

The average rate of change of f between 0 and 3 is:

m_{avg} = \dfrac{f\left(3\right) - f\left(0\right)}{3 - 0}

We have:

f\left(3\right) = 2^3 - 3^2

f\left(3\right) =8 - 9

f\left(3\right) = -1

And:

f\left(0\right) = 2^0 - 0^2

f\left(0\right) = 1 - 0

f\left(0\right) = 1

Therefore:

m_{avg} = \dfrac{-1 - 1}{3 - 0}

m_{avg} = -\dfrac{2}{3}