Help!!!!!!!!!!!!!!!!

Please help. Look at screenshot

Citrus2011 [14]

Answer:

a.

Increasing:

x < 0

x > 2

Decreasing:

0 < x < 2

b.

-1 < x < 2

x > 2

c.

x < -1

Step-by-step explanation:

a.

Function is increasing when it is going up as we go rightward

Function is decreasing when it is going down as we go rightward

We can see that as we move up (from negative infinity) until x = 0, the function is increasing. Also, as we go right from x = 2 towards positive infinity, the function is going up (increasing).

So,

Increasing:

x < 0

x > 2

The function is going down, or decreasing, at the in-between points of increasing, that is from 0 to 2, so that would be:

Decreasing:

0 < x < 2

b.

When we want where the function is greater than 0, we basically want the intervals at which the function is ABOVE the x-axis [ f(x) > 0 ].

Looking at the graph, it is

from -1 to 2 (x axis)

and 2 to positive infinity

We can write:

-1 < x < 2

x > 2

c.

Now we want when the function is less than 0, that is basically saying when the function is BELOW the x-axis.

This will be the other intervals than the ones we mentioned above in part (b).

Looking at the graph, we see that the graph is below the x-axis when it is less than -1, so we can write:

x < -1

4 0

4 years ago

Solve the following system of equations using matrices.

Alecsey [184]

It is convenient to let a calculator do the math.
.. (x, y) = (0, 3) . . . . . your 3rd selection

_____
I prefer using the RREF function to put the augmented matrix into reduced row-echelon form. If the equations are dependent or inconsistent, using that method tells you so without giving a matrix inversion error.