Let f(t)=t^(2)+5t+2.

Find a value of t such that the average rate of change of f(t) from 0 to t equals 11.

t=

rate of change=(tfinal-tinitial)/deltaT

11=t^2+5t+2-2 /t

11=t+5
t=6

To find the value of t such that the average rate of change of f(t) from 0 to t equals 11, we need to set up an equation. The average rate of change of a function is given by the formula:

Average rate of change = (f(t) - f(0)) / (t - 0)

In this case, we want the average rate of change to be 11. So we can set up the equation:

11 = (f(t) - f(0)) / (t - 0)

Let's substitute the function f(t) into the equation:

11 = ((t^2 + 5t + 2) - (0^2 + 5(0) + 2)) / t

Simplifying the equation:

11 = (t^2 + 5t + 2 - 2) / t
11 = (t^2 + 5t) / t

To solve this equation, we can multiply both sides by t:

11t = t^2 + 5t

Rearranging the equation to bring all terms to one side:

t^2 - 6t + 11t = 0
t^2 + 5t = 0

Factoring out t, we get:

t(t + 5) = 0

Setting each factor equal to zero:

t = 0 or t + 5 = 0

Solving for t, we find two possible values:

t = 0 or t = -5

Therefore, the value of t such that the average rate of change of f(t) from 0 to t equals 11 can be either t = 0 or t = -5.