How do you find local maximum value of f using the first and second derivative tests: #f(x) = x^5  5x + 5#?
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To find the local maximum value of ( f(x) = x^5  5x + 5 ) using the first and second derivative tests:

Find the critical points by setting the first derivative equal to zero and solving for ( x ). [ f'(x) = 5x^4  5 = 0 ] [ x^4  1 = 0 ] [ (x^2  1)(x^2 + 1) = 0 ] [ x = \pm 1 ]

Evaluate the second derivative at each critical point to determine concavity. [ f''(x) = 20x^3 ] At ( x = 1 ): ( f''(1) = 20 ), so the concavity is concave down. At ( x = 1 ): ( f''(1) = 20 ), so the concavity is concave up.

Apply the first derivative test:
 At ( x = 1 ): ( f'(1) = 5 ), indicating a local maximum.
 At ( x = 1 ): ( f'(1) = 5 ), indicating a local minimum.

Verify using the second derivative test:
 At ( x = 1 ): ( f''(1) = 20 < 0 ), confirming a local maximum.
 At ( x = 1 ): ( f''(1) = 20 > 0 ), confirming a local minimum.
Thus, the local maximum value of ( f(x) ) is at ( x = 1 ), with a maximum value of ( f(1) = 9 ).
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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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