how do i solve it algebraically? the teacher haven't taught derivatives
2007-12-06 12:04:56 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
minimum on what interval? in the interval (-∞,∞), there is no minimum because there's a vertical asymptote at x = 0. So just slightly left of 0, say x = -0.01, the 6x² part is positive, 0.00006, but the 2000/x is hugely negative, -200,000; and closer to x = 0 on the negative side, the fraction is even closer to -∞.
2007-12-06 12:14:07 · answer #1 · answered by Philo 7 · 1⤊ 0⤋
dy/dx = 12x - 2000/x^2 = 0, so x^3 = 500/3 and x = (500/3)^(1/3) is a critical number. Then d^2y/dx^2 = 12 + 4000/x^3, and for the critical value this is > 0, so our function is concave up there and it is indeed a minimum. Thus x is approximately 5.50321208149.
2007-12-06 20:19:22 · answer #2 · answered by Anonymous · 0⤊ 1⤋