Find all values of x that satisfy: 2x^3-x^2-8x+4>=0. I put ^ to tell you that it represents power and >= means equal to or greater than.
Find all values of x that satisfy `2x^3-x^2-8x+4 >= 0`
First we factor the left hand side:
`2x^3-x^2-8x+4` look at the terms two terms at a time
=`x^2(2x-1)-4(2x-1)` factoring out the common factor.
=`(x^2-4)(2x-1)` apply the distributive law
=`(x+2)(x-2)(2x-1)` factor the perfect square
So we now have `(x+2)(x-2)(2x-1)>=0`
Note that the function is zero at -2,2, and 1/2. We check the value of the function on the intervals x<-2; -2<x<1/2; 1/2<x<2; x>2
For x<-2 the function is negative (plug in a test value like -3)
For -2<x<1/2 the function is positive (plug in 0)
For 1/2<x<2 the function is negative (plug in 1)
For x>2the function is positive. (plug in 3)
We can also recognize that a cubic with leading coefficient greater than zero rises,falls, then rises and use the graph along with the zeros to find the intervals where the function is positive.
Thus your answer is `(-2 <= x <= 1/2)uu(x>2)`