To find the local extremum of f(x), we need to find the critical points of f(x), which occur when f'(x) = 0.
f'(x) = cos(x) - bsin(x)
Setting f'(x) = 0, we have:
cos(x) - bsin(x) = 0
We can rewrite this equation as:
cos(x) = bsin(x)
Dividing both sides by cos(x), we get:
1 = btan(x)
Therefore, we have b = tan(x).
So, the correct relationship to ensure a local extremum at x = k is b = tan(k).
For the inflection point of f(x), we need to find where f''(x) = 0 or where the second derivative changes sign.
f''(x) = -sin(x) - bcos(x)
Setting f''(x) = 0, we have:
-sin(x) - bcos(x) = 0
We can rewrite this as:
sin(x) = -bcos(x)
Dividing both sides by cos(x), we get:
tan(x) = -b
Therefore, the correct relationship to ensure an inflection point at x = k is b = -tan(k).
Suppose you are given the function f(x)=sinx+bcosx . Which of the following relationships will ensure that a local extremum occurs at x = k in the interval (0,π2) ? (1 point) Responses b=−2cotk b = − 2 cot k b=−cotk b = − cot k b=−tank b = − tan k b=cotk b = cot k b=tank b = tan k Question 2 Part B Which of the following relationships will ensure that an inflection point occurs at x = k in the interval (0,π2) ? (1 point) Responses b=−2cotk b = − 2 cot k b=−cotk b = − cot k b=−tank b = − tan k b=cotk b = cot k b=tank b = tan k Skip to navigation
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