How do you find the local maximum and minimum values of #f(x) = 2x^3 - 5x +1# in the the interval is (-3,3)?
graph{2x^3-5x+1 [-10, 10, -5, 5]}
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To find the local maximum and minimum values of ( f(x) = 2x^3 - 5x + 1 ) in the interval ((-3, 3)), follow these steps:
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Find the critical points by setting the derivative of ( f(x) ) equal to zero and solving for ( x ).
( f'(x) = 6x^2 - 5 )
Set ( f'(x) = 0 ) and solve for ( x ):
( 6x^2 - 5 = 0 )
( 6x^2 = 5 )
( x^2 = \frac{5}{6} )
( x = \pm \sqrt{\frac{5}{6}} )
The critical points are ( x = -\sqrt{\frac{5}{6}} ) and ( x = \sqrt{\frac{5}{6}} ).
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Evaluate ( f(x) ) at the critical points and at the endpoints of the interval ((-3, 3)).
( f(-\sqrt{\frac{5}{6}}) \approx 4.63 )
( f(\sqrt{\frac{5}{6}}) \approx -2.63 )
( f(-3) = -47 )
( f(3) = 52 )
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Compare the values obtained in step 2 to determine which one is the maximum and which one is the minimum within the given interval.
The local maximum value is ( f(3) = 52 ).
The local minimum value is ( f(-\sqrt{\frac{5}{6}}) \approx 4.63 ).
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When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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