Critical points is where the derivative (slope) is zero or does not exist. So to do this we have to find the derivative of our function:

So we apply chain rule:
=

Set our first derivative to zero and solve for x:
3(x^2 - 1) * 2x = 0
So we can see that (by plugging in) 0, -1 and 1 makes our solution true
So our critical value is x = 0, x = -1, x = 1
Answer:

Step-by-step explanation:
We need to find the value of 
Solving:
We know, 


a^0 = 1
so,

So, the value of 
Y=0 is the asymptote, and happy birthday
Answer:
-3r + 15 ---> answer
Step-by-step explanation:
r < 5
You are going to multiply both sides with 3. The reason being is that 3 is a positive number and the equality sign will not change if you use +3.
3r < 15
Now, subtract 15 from both sides, you will get this:
3r < 15
-15 -15
-------------
3r — 15 < 0
Lastly, using the Modulus function, we are going to add a negative sign to the content of our previous step because it's already negative.
So, -3r + 15 is the final solution if r < 5 in the given equation of l3r-15l
Answer:
33 1/3
Step-by-step explanation:
25/1 * 4/3 =
100/3 = 33.3... or 33 1/3