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?
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
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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Do your Own Homework
Good luck
Good luck