Answer:
B
Step-by-step explanation:
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!
Answer:
$49.28
Step-by-step explanation:
Multiply 44 by 1.12
If ƒ={(5, 1),(6, 2),(7, 3),(8, 1),(9, 7)}, then the range of ƒ is
yuradex [85]
1,2,3,7
I believe the range is just depending on your y values,and since an equation is not given your range should just be the y values of that set
Answer:
32x-24y-7
Step-by-step explanation:
Use the distributive property and multiply 4 by each term in the parentheses. 4 times 8x = 32x and 4 times -6y = -24y.
