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pashok25 [27]
3 years ago
10

state the domain, the range, and the intervals on which function is increasing, decreasing, or constant in interval notation

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
8 0

Answer:

  • domain (-∞, ∞)
  • range (-∞, 4]
  • increasing (-∞, 0)
  • decreasing (0, ∞)
  • constant (only at x=0, not on any interval)

Step-by-step explanation:

The graph is of the equation y = -x^2 +4. It is a polynomial of even degree, so has a domain of all real numbers: (-∞, ∞).

The vertical extent of the graph includes y=4 and all numbers less than that:

  range: (-∞, 4]

The graph is increasing to the left of its vertex at x=0, decreasing to the right.

  increasing (-∞, 0); decreasing (0, ∞)

There is no interval on which the function is constant. It has a horizontal tangent at x=0, but a single point does not constitute an interval.

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Students make 76.5 ounces of liquid soap for a craft fair they put the soap in 4.5 ounces bottles and sell each bottle for 5.59
liberstina [14]

Answer:

95.03

Step-by-step explanation: Divide 76.5 by 4.5

Now you have 17

Now, multiply 17 by 5.59.

Good job, these students are now successful coronavirus fighters.

8 0
3 years ago
I need help with this badly.
lesya [120]
Ok done. Thank to me :>

5 0
2 years ago
What are the steps for solving this
MatroZZZ [7]
Take -1-5 first and you'd get -6. then take 5×-6 and you'd get -30. finally take -30-40 and you'd get -70 as your answer
8 0
3 years ago
I think I know this much so far for A (but I could be wrong):
Eduardwww [97]
Part A. You have the correct first and second derivative.

---------------------------------------------------------------------

Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

-------------------------------------------------------------

Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out

To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h  ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0. 
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well. 
6 0
3 years ago
**Select all that apply**
Arte-miy333 [17]

Answer:

a and d

Step-by-step explanation:

8 0
3 years ago
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