Answer:
0/1 = 0
Step-by-step explanation:
0/1
0 ÷ 1 = 0
Answer:
(-5,3)
Step-by-step explanation:
<em><u>Method</u></em>
There are several ways you can do this, but one of the best ways is to use a graphing coordinate calendar. The most popular one is Desmos Graphing Calculator. I've attached an image of my work below using the system.
<u><em>Answer</em></u>
As you can see, both equations intersect at a certain coordinate point. This coordinate point is the point you need for the answer. The intersection is (-5,3).
I hope this helps. If you have any more questions, please feel free to post them and someone will be able to help you, whether it's myself or others. Please leave a like, rating, and if possible, Brainliest. Have a great day!
Answer:
A
Step-by-step explanation:
Answer:
1 is 6
Step-by-step explanation:
don't do my answer I'm probly wrong
(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.
