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find the relative extrema of the function: f(x)=x^4-6x^3+9x^2. Use the 2nd derivative test. list in ordered pr

list answers in ordered pairs

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1. The first derivative vanishes at x = 3 and x = 3/2. Since f"(3) > 0, there is a min at x = 3. Since f"(3/2) < 0, there is a max at x = 3/2. Find the ordered pairs yourself.
2. Relative extrema occur when f '(x) =0. Now f '(x) = 4x^3 - 18x^2 + 18x = 2x(2x^2 - 9x + 9) = 2x(2x-3)(x-3) so there are extrema when x=0, 3/2 and 3. The corresponding y-values are 0, (3/2)^2*((3/2)^2-6(3/2)+9) and 3^2*(3^2-6(3)+9) respectively = 0, 81/16, 0. Differentiating again gives f ''(x) = 12x^2 - 36x +18 = 6(2x^2 - 6x + 3) Evaluating this at the extrema gives f ''(0) = 6(3) =18, f ''(3/2) = 6(9/2 - 9 +3) = -9 f ''(3) = 6(18 - 18 +3) = 18 so by the 2nd derivative test we have local minima at (0,0) and (3,0) and a local maximum at (3/2, 81/16).