For the following questions solve for x, x∈R … a) (x+1)(x-2)(x-4)^2 ≥ 0 and b) -x^3 + 25x < 0

Answer 1

For (x+1)(x-2)(x-4)² ≥ 0

x+1=0, thus x=-1

x-2=0, thus x=2

(x-4)² is always greater or equal zero

Thus the answer is X≥2 and X≤-1

For -x³+25x<0, then x(25-x²)<0 , thus x(5-x)(5+x)<0

Thus, the answer for x(5-x)(5+x)<0 is -5<x<0 and x>5

Answer 2

We can use a wave curve method. We are going to first find out the critical point. Critical points are those points where your expression becomes zero or undefined.

f(x) = (x+1)(x-2)(x-4)²

f(x) become zero at -1,2 and 4. We are going to plot these on number line.

-------(-1)------------(2)--------------------(4)---------------

We see that these critical points have divided the number line into 4 parts. Now we are going to check the sign of f(x) in each parts. You can chose any point less than -1 to check sign of f(x) in that area. lets chose -2. f(x) = (-2+1)(-2-2)(-2-4)²= positive number. Similar do that others.

-------(-1)------------(2)--------------------(4)---------------

+ - + +

Since we want f(x) >= 0. x ∈ (-infinity,-1]∪[2, infinity)

Similarly, you can do that second part. If

I hope this helps.