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.
Answer:
B
Step-by-step explanation:
y≥-2x-2
put x=0,y=0
0≥-2
which is true.
so (0,0) lies on it.
graph is a solid line.
so B satisfies it.
Answer:
i dont get it please alaborate
Step-by-step explanation:
Answer:
30 m/s
Step-by-step explanation:
<em>p = mv</em>
<em>momentum</em><em> </em><em>=</em><em> </em><em>mass</em><em> </em><em>×</em><em> </em><em>velocity</em>
<em>here</em><em> </em><em>the</em><em> </em><em>momentum</em><em> </em><em>is</em><em> </em><em>60kgm</em><em>/</em><em>s</em>
<em>mass</em><em> </em><em>is</em><em> </em><em>2kg</em>
<em>substitute</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>equation</em>
<em>6</em><em>0</em><em> </em><em>=</em><em> </em><em>2v</em>
v= 30m/s
