Intervals that are increasing are 2, 4 and 7 ...
When you see the curve going upwards, it means it is increasing ...
Answer:
x=3
Step-by-step explanation:
In(x-1) = In(6) - In(x)
In(x-1) + In(x) = In(6)
In((x-1)x) = In(6)
In(x^2-x) = In(6)
x^2-x=6
x^2-x-6=0
x^2+2x-3x-6=0
x(x+2)-3(x+2)=0
(x-3)(x+2)=0
x=3 because x=-2 need to be cancel out
Answer:
For me i just guess lol i dont know the correct answer i just telling you the truth
Step-by-step explanation:
(a) When f is increasing the derivative of f is positive.
f'(x) = 15x^4 - 15x^2 > 0
15x^2(x^2 - 1)> 0
x^2 - 1 > 0 (The inequality doesn't flip sign since x^2 is positive)
x^2 > 1
Then f is increasing when x < -1 and x > 1.
(b) The f is concave upward when f''(x) > 0.
f''(x) = 60x^3 - 30x > 0
30x(2x^2 - 1) > 0
x(2x^2 - 1) > 0
x(x^2 - 1/2) > 0
x(x - 1/sqrt(2))(x + 1/sqrt(2)) > 0
There are four regions here. We will check if f''(x) > 0.
x < -1/sqrt(2): f''(-1) = -30 < 0
-1/sqrt(2) < x < 0: f''(-0.5) = 7.5 > 0
0 < x < 1/sqrt(2): f''(0.5) = -7.5 < 0
x > 1/sqrt(2): f''(1) = 30 > 0
Thus, f''(x) > 0 at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
Therefore, f is concave upward at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
(c) The horizontal tangents of f are at the points where f'(x) = 0
15x^2(x^2 - 1) = 0
x^2 = 1
x = -1 or x = 1
f(-1) = 3(-1)^5 - 5(-1)^3 + 2 = 4
f(1) = 3(1)^5 - 5(1)^3 + 2 = 0
Therefore, the tangent lines are y = 4 and y = 0.
Answer:
let missing number be x
63/21=9/x
or. 63x=9*21
or 63x=189
or x=189/63
or x=3
. missing number of this proportion=3
