Pre-Calc Help on Homework Please
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Pre-Calc Help on Homework Please

[From: ] [author: ] [Date: 12-04-01] [Hit: ]
Maximum and minimum values are correct,3.But this is not an absolute minimum,Therefore,......
Find the absolute maximum and minimum value. If they exist, over the interval indicated, if no interval is given, use the real line.

1. x^2 - 12x + 10.. I have minimum = -26 @ x = 6 w/ no max

2. 6x^2 - x^3 [-2,2]... I have max at 32 at x = 2 and minimum = 0 and x = 0

3. x^2 - 2/x.. not sure where to start?

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1. Correct

2. Maximum and minimum values are correct, but maximum (32) is at x = −2

3.

f(x) = x² − 2/x
f'(x) = 2x + 2/x²

Now we find values of x where f'(x) = 0
2x + 2/x² = 0
2x = −2/x²
2x³ = −2
x³ = −1
x = −1

f''(x) = 2 − 4/x³
f''(−1) = 2 − 4/−1 = 6 > 0

f(x) has LOCAL minimum at x = −1
f(−1) = (−1)2 − 2/−1 = 1 + 2 = 3 -----> local minimum

But this is not an absolute minimum, since f(1) = 1 − 2 = −1
Therefore, there are no absolute min/max for this function

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