How do you find the local extremas for #f(x)=2x + (5/x) #?
Local min is at
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To find the local extrema for ( f(x) = 2x + \frac{5}{x} ), you need to find the critical points where the derivative is either zero or undefined.

Find the derivative of ( f(x) ) using the power rule and the quotient rule. [ f'(x) = 2  \frac{5}{x^2} ]

Set the derivative equal to zero and solve for ( x ) to find critical points. [ 2  \frac{5}{x^2} = 0 ] [ 2x^2  5 = 0 ] [ 2x^2 = 5 ] [ x^2 = \frac{5}{2} ] [ x = \pm \sqrt{\frac{5}{2}} ]

Determine the nature of critical points using the second derivative test or by analyzing the behavior of the function around these points.

Since ( f'(x) = 2  \frac{5}{x^2} ), and ( f''(x) = \frac{10}{x^3} ), we can observe:
 At ( x = \sqrt{\frac{5}{2}} ), ( f'(x) > 0 ) and ( f''(x) > 0 ), indicating a local minimum.
 At ( x = \sqrt{\frac{5}{2}} ), ( f'(x) > 0 ) and ( f''(x) < 0 ), indicating a local maximum.
Therefore, the function has a local minimum at ( x = \sqrt{\frac{5}{2}} ) and a local maximum at ( x = \sqrt{\frac{5}{2}} ).
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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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