Part A. You have the correct first and second derivative.
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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.
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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.
(4 hundred + 0 tens + 5 ones) + (1 ten) = 4 hundred + 1 ten + 5 ones
... = 415
Using the critical point concept, it is found that a = -3 and b = 7.
<h3>What are the critical points of a function?</h3>
- The critical points of a function are the values of x for which:
In this problem, the function is:
Hence, the derivative is:
Then:
Since the critical point is at x = 2, we have that:
Then:
Critical point at (2,3) means that when 
, then:
You can learn more about the critical point concept at brainly.com/question/2256078
Answer:
<h3>Adjacent angles or Supplementary angles I am not sure
tho:)</h3>
Step-by-step explanation:
Answer: y = 1x+4
the related between the x and y is x plus one and y plus 4
Step-by-step explanation:
