Use the graph to find the local minimum and the local
maximum for the given function.
over the interval (-3, 0], the local minimum is
over the interval [0, 3], the local maximum is
over the interval [0, 57, the local minimum is

 The required answers are

(a) Local minimum over the interval (-3, 0] is x = -1.25 and the minimum value is -16.

(b) Local maximum over the interval [0, 3] is x = 1 and the maximum value is 4

and

(c) Local minimum over the interval [0, 5) is x = 2.5 and the minimum value is -3.

Step-by-step explanation:  We are given to use the graph to find the following local maximum and local minimum of the given function :

(a) Local minimum over the interval (-3, 0],

(b) Local maximum over the interval [0, 3]

and

(c) Local minimum over the interval [0, 5).

(a) From the graph, we see that over the interval (-3, 0], the minimum point of the curve is (-1.25, -16).

So, the local minimum of the given function over the interval (-3, 0] is x = -1.25 and the minimum value is -16.

(b) From the graph, we see that over the interval [3, 0], the maximum point of the curve is (1, 4).

So, the local maximum of the given function over the interval [3, 0] is x = 1 and the maximum value is 4.

(c) From the graph, we see that over the interval  [0, 5), the minimum point of the curve is (2.5, -3).

So, the local minimum of the given function over the interval [0, 5) is x = 2.5 and the minimum value is -3.

Thus, the required answers are

(a) Local minimum over the interval (-3, 0] is x = -1.25 and the minimum value is -16,

(b) Local maximum over the interval [0, 3] is x = 1 and the maximum value is 4

and

(c) Local minimum over the interval [0, 5) is x = 2.5 and the minimum value is -3.

1.0

step-by-step explanation:

it will be a cylinder. with a diameter of 8 and 4 single pieces high

step-by-step explanation: