* Solve for x, xER
A) (x+1) (x-2) (x-4)^2 > 0
B) -x^3+25<0
Thanks
A) Here is solution for - Wolfram|Alpha
1. Solve (x+1)(x-2)(x-4)^2 = 0
As you can see we have 3 roots here: $\displaystyle x_1=-1, x_2=2, x_3=4$. It means function $\displaystyle y=(x+1)(x-2)(x-4)^2$ changes it's sign in this points.
2. Find sign of y in following intervals by selecting random value and evaluating y: (-inf, -1), (-1, 2), (2, 4) and (4,+inf).
3. Select intervals where y > 0
Helolo, agent2421!
Note that $\displaystyle (x-4)^2$ is never negative.A) Solve for $\displaystyle x\!:\;\;(x+1) (x-2) (x-4)^2 \:\geq\:0$
So we are concerned with $\displaystyle (x+1)$ and $\displaystyle (x-2)$ only.
The product is greater than or equal to zero in two cases:
. . (1) Both factors are positive
. . (2) Both factors are negative
Case 1: .$\displaystyle \begin{array}{ccccccc}x+1 & \geq & 0 & \Rightarrow & x & \geq & \text{-}1 \\ x-2 &\geq & 0 & \Rightarrow & x & \geq & 2 \end{array} \quad\Rightarrow\quad x \:\geq\:2$
Case 2: .$\displaystyle \begin{array}{ccccccc}x + 1 & \leq & 0 & \Rightarrow & x & \leq & \text{-}1 \\ x-2 & \leq & 0 & \Rightarrow & x & \leq & 2 \end{array}\quad\Rightarrow\quad x \:\leq \:\text{-}1$
Therefore: . $\displaystyle (x \:\leq\: \text{-}1) \:\vee \:(x \:\geq\: 2)\quad \hdots \text{or: }\;(-\infty,\:\text{-}1] \:\cup\:[2,\,\infty)
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