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Find the focus, directrix, and vertex for the parabola x² = -16y.pdf
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Find the extrema for the following function on their given interval..pdf
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Find the extrema for the following function on their given interval. Label extreme values as either absolute, local, or both. Solution domain is x<=3 f\'(x) = 2x sqrt(3-x) - x^2 / 2* sqrt(3-x) f\'(x) = 0 so 2x * (3-x) - x^2 /2 = 0 => +6x -3x^2 -0.5 x =0 so x= 0 0r x= 5.5/3 = 1.83 f\'\'(x) = 2* sqrt(3-x) - x/sqrt(3-x) - x / sqrt(3-x) - x^2/4(sqrt(3-x)) at x=0 f\'\' is > 0 so minima at 0 and at x =1.83 f\'\' < 0 so minima at x=1.83.
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Find the extrema for the following function on their given interval..pdf
- Find the extrema for the following function on their given interval. Label extreme values as either absolute, local, or both. Solution domain is x<=3 f'(x) = 2x sqrt(3-x) - x^2 / 2* sqrt(3-x) f'(x) = 0 so 2x * (3-x) - x^2 /2 = 0 => +6x -3x^2 -0.5 x =0 so x= 0 0r x= 5.5/3 = 1.83 f''(x) = 2* sqrt(3-x) - x/sqrt(3-x) - x / sqrt(3-x) - x^2/4(sqrt(3-x)) at x=0 f'' is > 0 so minima at 0 and at x =1.83 f'' < 0 so minima at x=1.83
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