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:
[-2, ∞]
Step-by-step explanation:
15 - √(x+2)
domain is any value as long as (x+2) is not-negative, since √ of a negative number has no Real solution.
x+2 ≥ 0 ⇒ x ≥ -2
Answer:
Trufant is 28 while his younger brother(Yolensi) is 19
The answer is A because 24 3/8 inches is roughly 2 feet, 100/2 is 50, and since there is a small 3/8 value, the answer is 49
Express √3 + i in polar form:
|√3 + i| = √((√3)² + 1²) = √4 = 2
arg(√3 + i) = arctan(1/√3) = π/6
Then
√3 + i = 2 (cos(π/6) + i sin(π/6))
By DeMoivre's theorem,
(√3 + i)³ = (2 (cos(π/6) + i sin(π/6)))³
… = 2³ (cos(3 • π/6) + i sin(3 • π/6))
… = 8 (cos(π/2) + i sin(π/2))
… = 8i