How you could show five less than a number using an algebraic expression
1 answer:
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You are correct. The answer is choice D
The only way for g(x) to be differentiable at x = 0 is for two things to happen
(1) g(x) is continuous at x = 0
(2) g ' (x) is continuous at x = 0
To satisfy property (1) above, the value of b must be 1. This can be found by plugging x = 0 into each piece of the piecewise function and solving for b.
So the piecewise function becomes
after plugging in b = 1
--------------------------------
Now differentiate each piece with respect to x to get
The first piece of g ' (x) is always going to be equal to 1. The second piece is equal to zero when x = 0
Because -sin(x) = -sin(0) = 0
So there's this disconnect on g ' (x) meaning its not continuous
Therefore, the value b = 1 will not work.
So there are no values of b that work to satisfy property (1) and property (2) mentioned at the top.
The only way for g(x) to be differentiable at x = 0 is for two things to happen
(1) g(x) is continuous at x = 0
(2) g ' (x) is continuous at x = 0
To satisfy property (1) above, the value of b must be 1. This can be found by plugging x = 0 into each piece of the piecewise function and solving for b.
So the piecewise function becomes
after plugging in b = 1
--------------------------------
Now differentiate each piece with respect to x to get
The first piece of g ' (x) is always going to be equal to 1. The second piece is equal to zero when x = 0
Because -sin(x) = -sin(0) = 0
So there's this disconnect on g ' (x) meaning its not continuous
Therefore, the value b = 1 will not work.
So there are no values of b that work to satisfy property (1) and property (2) mentioned at the top.