x^5+3x^3 + x -1
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I think this is y = x^5 + 3x^3 + x for -1 < x < 2
dy/dx = 5x^4 + 9x^2 + 1 = 0 ==> x^2 = [-9 ± √(81 - 20)] / 10 < 0
The roots of the derivative are imaginary. So we'll check some points.
For x = -1: y = -1 - 3 - 1 = -5
For x = 0: y = 0
For x = 1: y = 1 + 3 + 1 = 5
For x = 2: y = 32 + 24 + 2 = 58
Over the interval, the minimum is -5+ and the maximum is 58-. (Answer)
dy/dx = 5x^4 + 9x^2 + 1 = 0 ==> x^2 = [-9 ± √(81 - 20)] / 10 < 0
The roots of the derivative are imaginary. So we'll check some points.
For x = -1: y = -1 - 3 - 1 = -5
For x = 0: y = 0
For x = 1: y = 1 + 3 + 1 = 5
For x = 2: y = 32 + 24 + 2 = 58
Over the interval, the minimum is -5+ and the maximum is 58-. (Answer)
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5x^4 + 9x^2 + 1 = 0 is a quadratic in x^2. I used the quadratic formula.
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