Question

Given the function h(x)=-x^2+4x+11, determine the average rate of change of the function over the interval 0≤x≤6.

Answer 1

The function is given as,

`h(x)\text{ = -x}^2+4x+11\text{ over the interval 0 }\leq\text{ x }\leq\text{ 6.}`

The average rate of change of the function over the interval x [a,b] is given by,

`(h(b)-h(a))/(b-a)`

The average rate of change of the function over the interval,

`\begin{equation*} \text{ 0 }\leq\text{ x }\leq\text{ 6.} \end{equation*}`

is given by,

`\frac{h(6)\text{ - h\lparen0\rparen}}{6-0}\text{ = }(-1+11)/(6)\text{ = }(10)/(6)\text{ = }(5)/(3)`

Thus the average rate of change is 5/3 or 1.67.