Question

27x^4-3x^2>=0 решите неравенство

Answer 1

Решение задания приложено

Answer 2

хорошо что без графика

Answer 3

To solve the inequality $27x^4 - 3x^2 \geq 0$, we first factor it to find its critical points and then determine the intervals where the inequality is satisfied.

Step 1: Factor the inequality: $27x^4 - 3x^2 = 3x^2(9x^2 - 1) = 3x^2(3x + 1)(3x - 1)$

Step 2: Find critical points (where the expression equals zero): Set each factor equal to zero and solve for $x$:

1. $3x^2 = 0 \Rightarrow x = 0$
2. $3x + 1 = 0 \Rightarrow 3x = -1 \Rightarrow x = -1/3$
3. $3x - 1 = 0 \Rightarrow 3x = 1 \Rightarrow x = 1/3$

Step 3: Use test points to determine the intervals where the inequality is satisfied: We now need to find the intervals on the number line where the expression $27x^4 - 3x^2$ is greater than or equal to zero. We will choose test points from each interval and evaluate the expression to determine its sign.

Choose a test point $x = -2$ (a value less than $-1/3$): $27(-2)^4 - 3(-2)^2 = 27(16) - 3(4) = 432 - 12 = 420$ (positive)

Choose a test point $x = 0$ (a value between $-1/3$ and $1/3$): $27(0)^4 - 3(0)^2 = 0$ (zero)

Choose a test point $x = 1/2$ (a value between $0$ and $1/3$): $27(1/2)^4 - 3(1/2)^2 = 27(1/16) - 3(1/4) = 27/16 - 3/4 = 27/16 - 12/16 = 15/16$ (positive)

Choose a test point $x = 2$ (a value greater than $1/3$): $27(2)^4 - 3(2)^2 = 27(16) - 3(4) = 432 - 12 = 420$ (positive)

Step 4: Analyze the results and state the solution: From our evaluations, we can see that the inequality $27x^4 - 3x^2 \geq 0$ is satisfied for $x \leq -1/3$ and $x \geq 1/3$. The interval notation for the solution is: $x \in (-\infty, -1/3] \cup [1/3, \infty)$