(01.02 MC)

The distance, d(t), in feet, a bug has traveled is shown in the graph.

A coordinate plane with a function d of t consisting of a line starting at 0 comma 1 rising to 1 comma 2, a horizontal line from 1 comma 2 to 2 comma 2, a line falling from 2 comma 2 to 3 comma 1, and a line falling from 3 comma 1 to 6 comma 0.

Estimate the rate of change in the distance of the bug at time t = 4 seconds.

negative one third ft/s
the limit as t approaches 4 of the function d of t feet per second
d(4) ft/s
0.7 ft/s

Respuesta :

When we have a given function f(x), the rate of change in a given value x₀ is given by:

[tex]\frac{df(x_0)}{dx} = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}[/tex]

We will find that the rate of change at t = 4s is:

r = -1/3

Now we can't do this if we do not have the function and we only have a text description of the graph, but what we can do is find the average rate of change.

For a function f(x), the average rate of change in an interval (a, b) such that a < b is given by:

[tex]r = \frac{f(b) - f(a)}{b - a}[/tex]

Something really nice is that the average rate of change is equal to the exact rate of change if f(x) is a linear equation.

If we want to find the rate of change at t = 4, then we need to find the smallest interval that contains t = 4.

Here we know that the graph passes through the points:

(3,1) to (6, 0)

Because of the statement "a line falling from (3, 1) to (6, 0)" we know that in this segment we have a line, which implies that the average rate of change will be equal to the exact rate of change.

Using the equation of the average rate of change we get:

[tex]r = \frac{d(6) - d(3)}{6 - 3} = \frac{0 - 1}{3} = - 1/3[/tex]

Then the rate of change in the distance at the time t = 4s is -1/3.

If you want to learn more, you can read:

https://brainly.com/question/18904995