Critical values are values where f'(x)=0 and the bounds of a function. Thus, let's solve for f'(x)!
f(x)=2x^3-3x^2+3x+8
f'(x)=6x^2-6x+3
Now let's set f'(x)=0
0=6x^2-6x+3
0=2x^2-2x+1
As it turns out, 2x^2-2x+1 isn't factorable!
This saves me some time because this means there are no critical numbers!
I don't think that is even possible :|
I was taught that dividing by zero is undefined. Dividing 1 by 0 means 1/0 which makes no sense at all.
But all I'm saying is that we can't divide by zero.
Try this 0/0 does it make sense to you? Nah that is indeterminate.
If you mean the gradient then it would be -3/2 because if you multiply the 2 together they equal minus 1. That’s how u figure out if 2 lines are perpendicular
Answer:
Step-by-step explanation: