Solve for x 8x+1>−31
1 answer:
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Answer:
No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R
Step-by-step explanation:
Assuming: the function is in [0,1]
And rewriting it for the sake of clarity:
Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer
1) A function is considered to be differentiable if, and only if both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:
2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]
The limit to the left
g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)
3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R
Because this is the same as to calculate the limit from the left and right side, of g(x).
This is what the Bilateral Theorem says: