(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:
X=2
Step-by-step explanation:
Answer:
0.2745%
Step-by-step explanation:
69.95-42.50=27.45
27.45×100=27.45%
X=2
inscribed angle - 3(2)+4=10
arc - 10(2)=20
the Arc is two times the inscribed angle
Answer:
infinitely many solutions
Step-by-step explanation:
solve by substitution:
solve equation 1 for y: - y = 7 - 4x; divide by -1 to simplify into y = -7 + 4x
Substitute - 7 + 4x into the second equation to obtain:
3(-7 + 4x) - 12x = -21, which yields:
-21 + 12x - 12x = -21, which simplifies 0 = 0
a result of 0 = 0 means infinite solutions such as:
(2,1), (3, 5), (4, 9), (5, 13), (6, 17)....
