We first begin by drawing the graph of the given function:

`f(x) (=y) =1/4*cos(4*pi*x)-3 `

The graph is below attached.

Then we remember what the average rate of change of a function `f(x) ` between `x_1 ` and `x_2 ` means

`(Delta(f(x)))/(Delta(x)) = (f(x_1)-f(x_2))/(x_1-x_2)` (1)

Graphically this is the slope...

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We first begin by drawing the graph of the given function:

`f(x) (=y) =1/4*cos(4*pi*x)-3 `

The graph is below attached.

Then we remember what the average rate of change of a function `f(x) ` between `x_1 ` and `x_2 ` means

`(Delta(f(x)))/(Delta(x)) = (f(x_1)-f(x_2))/(x_1-x_2)` (1)

Graphically this is the slope of the line that connects the points `f(x_1) ` and `f(x_2) ` (line 1 in the figure)

For example the rate of change between `x_1 =-0.1 ` and `x_2 =+0.2 ` is positive since the slope of the line 1 in figure is positive.

Now, the instantaneous rate of change of function around point `x_1 ` is when `x_2 =x_1+h ` and `h ->0 `.

This graphically means that the line drawn between points `x_1 ` and `x_2=x_1+h ` in graph becomes tangent to the graph of function `f(x) ` at point `x_1 ` .

To find two points where the instantaneous rate of change of function is zero, we need to find two horizontal tangent lines to the graph. These are lines 2 and 3 in the figure. The corresponding values of `x ` are `x =0 ` and `x=0.25 `

**Thus two points where the rate of the change of the given function is zero are `x =0 ` and `x =0.25 ` .**

For positive values of rate of change the tangent lines need to have a positive slope (need to increase). These are lines 4 and 5 in the figures.

**Thus two points where the rate of change of the given function is positive are `x=0.3 ` and `x=0.4 ` .**

For negative values of rate of change the tangent lines need to have negative slope (need to decrease). These are the lines 6 and 7 in the figure.

**Thus two points where the rate of change of the given function is negative are `x =-0.3 ` and `x=0.55 ` **

**Further Reading**