How do you find the exact relative maximum and minimum of the polynomial function of #f(x) = x^3 + x + 4/x#?
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To find the exact relative maximum and minimum of the polynomial function ( f(x) = x^3 + x + \frac{4}{x} ), follow these steps:
 Find the derivative of the function ( f'(x) ).
 Set ( f'(x) ) equal to zero and solve for ( x ) to find critical points.
 Determine the second derivative ( f''(x) ).
 Evaluate ( f''(x) ) at each critical point.
 If ( f''(x) > 0 ), the function has a relative minimum at that point.
 If ( f''(x) < 0 ), the function has a relative maximum at that point.
Let's solve it step by step:

( f(x) = x^3 + x + \frac{4}{x} ) ( f'(x) = 3x^2 + 1  \frac{4}{x^2} )

Set ( f'(x) = 0 ): ( 3x^2 + 1  \frac{4}{x^2} = 0 ) ( 3x^4 + x^2  4 = 0 ) ( (x^2 + 4)(3x^2  1) = 0 ) ( x^2 = 4 ) or ( x^2 = \frac{1}{3} ) Since ( x^2 = 4 ) has no real solutions, we consider ( x^2 = \frac{1}{3} ). ( x = \pm \frac{1}{\sqrt{3}} )

Find ( f''(x) ): ( f''(x) = 6x + \frac{8}{x^3} )

Evaluate ( f''(x) ) at critical points: ( f''\left(\frac{1}{\sqrt{3}}\right) = 6\left(\frac{1}{\sqrt{3}}\right) + \frac{8}{\left(\frac{1}{\sqrt{3}}\right)^3} ) ( f''\left(\frac{1}{\sqrt{3}}\right) > 0 ) (positive), so relative minimum exists at ( x = \frac{1}{\sqrt{3}} ).
( f''\left(\frac{1}{\sqrt{3}}\right) = 6\left(\frac{1}{\sqrt{3}}\right) + \frac{8}{\left(\frac{1}{\sqrt{3}}\right)^3} ) ( f''\left(\frac{1}{\sqrt{3}}\right) < 0 ) (negative), so relative maximum exists at ( x = \frac{1}{\sqrt{3}} ).
Therefore, the exact relative maximum occurs at ( x = \frac{1}{\sqrt{3}} ), and the exact relative minimum occurs at ( x = \frac{1}{\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.
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