check out the difference:
d(x) = (4x^3+14x^2-2x-3)-(2x^2-x) = 4x^3+12x^2-3x-3
where is that equal to zero? Hard to factor, so getting out your handy polynomial solver, we find that
x = -3.1622, -0.4126, 0.5748
Now, since d(x) is a cubic with positive leading coefficient, we know that it is negative to the left of -3.1622. Since it changes sign every time it passes through a root, we have
d(x) > 0 on the set (-3.1622,-0.4126) U (0.5748,∞)
For what values of x does the graph of the function y=4x^3+14x^2-2x-3 lie above the function y=2x^2-x?
3 answers
d(x) = 4 x³ + 14 x² - 2 x - 3 - ( 2 x² - x ) = 4 x³ + 14 x² - 2 x - 3 - 2 x² + x =
4 x³ + 12 x² - x - 3
Factorisation:
d(x) = 4 x³ + 12 x² - x - 3 = ( x + 3 ) ( 2 x + 1 ) ( 2 x - 1 )
Solutions x = - 3 , x = - 1 / 2 and x = 1 / 2
- 3 < x < - 1 / 2 OR x > 1 / 2
( - 3 , - 1 / 2 ) ∪ ( 1 / 2 ,∞)
4 x³ + 12 x² - x - 3
Factorisation:
d(x) = 4 x³ + 12 x² - x - 3 = ( x + 3 ) ( 2 x + 1 ) ( 2 x - 1 )
Solutions x = - 3 , x = - 1 / 2 and x = 1 / 2
- 3 < x < - 1 / 2 OR x > 1 / 2
( - 3 , - 1 / 2 ) ∪ ( 1 / 2 ,∞)
Nice catch, Bosnian. Guess I should have watched those minus signs ...