What are the global and local extrema of #f(x) = x^39x+3 # ?
There are no global extrema.
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To find the global and local extrema of ( f(x) = x^3  9x + 3 ):

Take the derivative of ( f(x) ) to find critical points. ( f'(x) = 3x^2  9 )

Set the derivative equal to zero and solve for ( x ): ( 3x^2  9 = 0 ) ( x^2  3 = 0 ) ( x^2 = 3 ) ( x = \pm \sqrt{3} )

Evaluate the second derivative to determine the nature of these critical points. ( f''(x) = 6x )

Evaluate the second derivative at the critical points: ( f''(\sqrt{3}) = 6\sqrt{3} ) (positive, so minima) ( f''(\sqrt{3}) = 6\sqrt{3} ) (negative, so maxima)

Substitute critical points into the original function to find the corresponding function values: ( f(\sqrt{3}) = (\sqrt{3})^3  9(\sqrt{3}) + 3 ) ( f(\sqrt{3}) = (\sqrt{3})^3  9(\sqrt{3}) + 3 )

Compare the values to find global and local extrema.
Global minimum: ( f(\sqrt{3}) ) Global maximum: ( f(\sqrt{3}) )
Local minimum: ( f(\sqrt{3}) ) Local maximum: ( f(\sqrt{3}) )
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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