if f(x)=x(x+3)(x-1) give the value of x where f(x)>0?

Question

I have to express it in interval notation

Answer 1

I'll walk through the question:

Given that f(x) = x (x+3) (x-1), find x where f(x) > 0

Substitute:

x (x+3) (x-1) > 0

Which values of x will make x (x+3) (x-1) = 0?

-3, 0, and 1

Number line test:

Will values of x less than -3 make the inequality true?
-4 (-4+3) (-4-1) is not greater than zero

Between -3 and 0?
-1 (-1+3) (-1-1) > 0 ***so -3 < x < 0***

Between 0 and 1?
0.5 (0.5+3) (0.5-1) is not greater than zero

Greater than 1?
2 (2+3) (2-1) > 0 ***so x > 1***

Answer:

-3 < x < 0 or x > 1

Answer 2

Find the zeros of the function (set each factor =0 and solve) and test the values between these zeros. For this function, your zeros are at -3, 0 and 1. So test a value smaller than -3, one that is between -3 and 0, one that is between 0 and 1 and one larger than 1, you'll find the function is greater than 0 between -3 and 0 and also larger than 1.

Answer 3

3x = 0 3x / 3 = 0 / 3 x = 0 (you opt to locate x and because there's a coefficent of three in front of x.. you opt to do away with it via divinding the two components.) now which you already know x=0 sub it right into a million + x + x^2 a million + (0) + (0)^2 a million + 0 + 0 = a million hence the respond is B

Answer 4

the answer is x=1. You can get this because if you look at the factors the three roots are X= 0, 1, -3. Therefore the only one greater than (f(x)>0) is 1 which you get from f(x)=(x-1).

Answer 5

f(x)>0 where
-31

Solve by determining where f(x) = 0 and determine where the function is positive between the intervals.

Answer 6

zero points: x = -3,0,1
f(x)>0 when x>1 or -3

Answer 7

I think it is everything over one