Question

g(t)= -(t-1)^2 +5g(t)=−(t−1) 2 +5g, (, t, ), equals, minus, (, t, minus, 1, ), squared, plus, 5 What is the average rate of change of ggg over the interval -4\leq t\leq 5−4≤t≤5minus, 4, is less than or equal to, t, is less than or equal to, 5?

Answer 1

Average rate of change is 1.

Explanation:

The average rate of change of function
`f` over the intervals
`a\leqslant x\leqslant b` is given by the formula,

`A\left(x\right)=(f\left(b\right)-f\left(a\right))/(b-a)`.

Rewriting the formula according to given data,

`A\left(t\right)=(g\left(b\right)-g\left(a\right))/(b-a)`.

Given the interval as
`-4\leqslant t\leqslant 5`. Hence value of a and b is a = -4 and b = 5.

Now calculate
`g\left(b\right)` and
`g\left(a\right)` as follows.

Calculation for
`g\left(a\right)`,

`g\left(t\right)=-\left(t-1\right)^2+5`

Replace t as a,

`g\left(a\right)=-\left(a-1\right)^2+5`

Substituting the value,

`g\left(-4\right)=-\left(-4-1\right)^2+5`

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

`g\left(-4\right)=-25+5`

`g\left(-4\right)=-20`

Calculation for
`g\left(b\right)`,

`g\left(b\right)=-\left(b-1\right)^2+5`

Substituting the value,

`g\left(5\right)=-\left(5-1\right)^2+5`

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

`g\left(5\right)=-16+5`

`g\left(5\right)=-11`

Now substituting the value,

`A\left(t\right)=(-11-\left(-20\right))/(5-\left(-4\right))`.

Simplifying,

`A\left(t\right)=(-11+20)/(5+4)`.

`A\left(t\right)=(9)/(9)`.

`A\left(t\right)=1`.

Hence average rate of change of the given function
`g\left(t\right)=-\left(t-1\right)^2+5` over the given interval
`-4\leqslant t\leqslant 5` is 1.