Respuesta :

altavistard

Answer:

(0, 2)

Step-by-step explanation:

The graph of g(t)= -(t-1)^2+5 is an inverted parabola with vertex at (1, 5).

Making a table of t and g values would be helpful here:

 t      g(t) = -(t - 1)^2 + 5

------ -----

  2     4

  0     4

 -1      1

  1      5

We're looking for an interval on which the average rate of change is zero.

Note that this is the case on the interval (2, 4); g(0) = g(2) = 4, so the change in g is 4 - 4, or zero (0).

deepakrai138

The average rate of change of [tex]g(t)=-(t-1)^2+5[/tex] is 0 in interval [tex]-1\leq t\leq 3[/tex].

Given,

[tex]g(t)=-(t-1)^2+5\\[/tex].

We have to find the interval in which [tex]g(t)=-(t-1)^2+5\\[/tex] have an average rate of change of zero.

We know that, the function [tex]f(x)[/tex] will have average range of 0 when [tex]f(b)=f(a)[/tex].

Now we calculate g(1), g(2),g(3) and g(-1),

[tex]g(1)=-(1-1)^2+5\\g(1)=5[/tex]

[tex]g(2)=-(2-1)^2+5\\g(2)=-1+5\\g(2)=4[/tex]

[tex]g(3)=-(3-1)^2+5\\g(3)=-4+5\\g(3)=1[/tex]

[tex]g(-1)=-(-1-1)^2+5\\g(-1)=-4+5\\g(-1)=1[/tex]

Since,

[tex]g(3)=g(-1)=1[/tex]  so the function [tex]g(t)=-(t-1)^2+5\\[/tex] has an average rate of zero at [tex]-1\leq t\leq 3[/tex].

For more details follow the link:

https://brainly.com/question/2530409