32*10=320 Hope you like and rate
To solve this, you have to know that the first derivative of a function is its slope. When an interval is increasing, it has a positive slope. Thus, we are trying to solve for when the first derivative of a function is positive/negative.
f(x)=2x^3+6x^2-18x+2
f'(x)=6x^2+12x-18
f'(x)=6(x^2+2x-3)
f'(x)=6(x+3)(x-1)
So the zeroes of f'(x) are at x=1, x=-3
Because there is no multiplicity, when the function passes a zero, he y value is changing signs.
Since f'(0)=-18, intervals -3<x<1 is decreasing(because -3<0<1)
Thus, every other portion of the graph is increasing.
Therefore, you get:
Increasing: (negative infinite, -3), (1, infinite)
Decreasing:(-3,1)
Answer:
Use the formula for direct variation
Step-by-step explanation:
Jk=8x+6
J=(8x+6)/k
kL=6x+20
L=(6x+20)/k
JL=(8x+6)/k * (6x+20)/k
=(48x²+196x+120)/k²
I think the answer is C.
First start by factoring, you should get (2n-3)(n-2)=0
Then start to solve for n.
2n-3=0
n-2=0
n=2, n= 3/2
I hope that helps.
