2017年4月24日 · See below. For my approach, I will be using a graphical interpretation. You can rewrite the equation as x^3 - x - 1 = 0 as the first step. Then graph the following: f(x) = x^3 - x - 1. graph{x^3 - x - 1 [-10, 10, -5, 5]} Click on where the graph intersects the x-axis. This point should be (1.325, 0). Therefore, the answer is x = 1.325.

2015年4月8日 · x^2 - 3x = 0 x*(x-3) = 0 (x was a common factor to both the terms) In general, if a*b = 0, then either a ...

2025年2月27日 · 7 – 4 + 3 x 0 + 1 = ? The author is challenging what the average person knows. The correct answer is, of course, 4. So many people out there insist that it's 1. After all, anything times 0 is 0. I've even seen some people claim the answer is 7, but I suspect that's a matter of glossing over the details and mistaking the multiplication for ...

2016年3月13日 · How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#?

2016年11月30日 · f(x)≥0 for 1≤x≤3 with the exclusion of x=2 where it is undefined. The denominator is always positive, so the sign of f(x) is determined by the numerator. The numerator is positive when both factors are positive or both are negative.

2017年6月13日 · Solution for (x+3)/(x-1) >= 0 is the region (-oo,-3]uu[1.oo) i.e. {x:x<=-3 or x>=1} The sign of (x+3)/(x-1) depends on the values (x+3) and (x-1) take at any point. It is apparent that change of sign will appear around x=-3. Observe before that (x+3) is negative and after that it is positive. Similarly, before x=1, (x-1) is negative and after that it is positive. Now, leave the …

2015年10月7日 · Use the fact that if absx < 1, then abs(x^3) < abs x. Given epsilon > 0, choose delta = min{1, epsilon} Now if 0 < abs(x-0) < delta then . . . etc. to show abs(x^3-0) < epsilon.

2018年6月3日 · The roots are -1,-1,2 It is easy to see by inspection that x = -1 satisfies the equation : (-1)^3-3times(-1)-2 = -1+3-2=0 To find the other roots let us rewrite x^3-3x-2 keeping in mind that x+1 is a factor: x^3-3x-2 = x^3+x^2-x^2-x-2x-2 qquadqquad = x^2(x+1)-x(x+1)-2(x+1) qquadqquad = (x+1)(x^2-x-2) qquadqquad = (x+1)(x^2+x-2x-2) qquadqquad = (x+1){x(x+1) …

2016年7月26日 · 5018 views around the world You can reuse this answer ...

2017年1月2日 · -3 < x < 4. Find the critical points. Equate the "factors" to 0 to get the critical points. Use 0 as a critical point also. x - 4 = 0" "=>" "x = 4 x + 3 = 0" "=>" "x = "-"3 The critical points are {"-"3, 0, 4}. It is only at these points that the sign of (x-3)(x+4) may change. Now, pick a number that lies in each region (in-between/on either side of these critical points), plug it into …