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Sagot :
f'(x)>0 for all x>0. Therefore f(x) is strictly increasing on the inverval (0,∞).
How to obtain the maximum value of a function?
To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.
f(x) = max{0,x} then f(x)=0 if x<0 and f(x)=x if x≥0.
f(x)=7x-2xlnx
When x tends to -∞, f is the constant function zero, therefore limit n tends to infinity f(x) = 0.
When x tends to ∞, f(x)=x and x grows indefinitely. Thus limit n tends to infinity f(x) = infinity.
f is differentiable if x≠0. If x>0, f'(x)=1 (the derivative of f(x) = x and if x<0, f'(0)=0 (the derivative of the constant zero).
In x=0, the right-hand derivative is 1, but the left-hand derivative is 0, hence f'(0) does not exist,
f'(x)>0 for all x>0. Therefore f(x) is strictly increasing on the inverval (0,∞).
Learn more about maxima of a function here:
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