1. ## Math Problem Help

* Solve for x, xER

A) (x+1) (x-2) (x-4)^2 >
0

B) -x^3+25<0

Thanks

2. Originally Posted by agent2421
* Solve for x, xER

A) (x+1) (x-2) (x-4)^2 > 0

B) -x^3+25<0

Thanks
A) Answer this question by first drawing a graph of y = (x+1) (x-2) (x-4)^2 (the x-intercepts are all you'll need to show on the graph).

B) -x^3 < -25 => x^3 > 25 and the last step should be obvious.

3. 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

4. Helolo, agent2421!

A) Solve for $\displaystyle x\!:\;\;(x+1) (x-2) (x-4)^2 \:\geq\:0$
Note that $\displaystyle (x-4)^2$ is never negative.

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: . $\displaystyle (x \:\leq\: \text{-}1) \:\vee \:(x \:\geq\: 2)\quad \hdots \text{or: }\;(-\infty,\:\text{-}1] \:\cup\:[2,\,\infty)$

5. so my final answer would be... (x< 1) (x>2) ?

Like if I put that on my answer sheet, it is the correct one?

6. I think you did typo.

The answer is x<=-1 and x>=2