The solution below corresponds to an inequality involving a quartic function.
Write a possible quartic polynomial inequality that matches the solution. Prove that your inequality matches the solution.
x < 4 and x > 7
if r is a root, then (x-r) is a factor
(x-4)(x-7) = x^2 - 11x + 28
x^2 - 11x + 28 > 0
the parabola crosses the x-axis at 4 and 7
... function values outside of the roots are greater than zero
but i need a quartic equation
you know that x^4 > 0 has solutions
x < 0 or x > 0
So, since 11/2 is midway between 4 and 7, just shift right and you have
4-11/2 = 3/2
7-11/2 = 3/2
so,
(x - 11/2)^4 > (3/2)^4
see
http://www.wolframalpha.com/input/?i=(x+-+11%2F2)%5E4+%3E+(3%2F2)%5E4
and scroll down to where it shows the solution set
Thanks a lot
To write a quartic polynomial inequality that matches the solution x < 4 and x > 7, we can consider the range of values that satisfy these conditions. Since x < 4 implies that x is less than 4, and x > 7 implies that x is greater than 7, we can conclude that x must be between these two values, i.e., x ∈ (4, 7).
A quartic polynomial inequality that matches this solution is:
(x - 4)(x - 7)(x - α)(x - β) < 0
Here, α and β are two arbitrary real numbers that lie between 4 and 7, and the quartic function is defined as f(x) = (x - 4)(x - 7)(x - α)(x - β).
To prove that the above inequality matches the solution x < 4 and x > 7, we need to verify that it satisfies the given conditions.
1. For x < 4:
- If we choose α and β such that 4 < α < β < 7, then when x < 4:
- (x - 4) < 0
- (x - 7) < 0
- (x - α) < 0
- (x - β) < 0
- The product of these factors is negative, which satisfies the condition for x < 4.
2. For x > 7:
- If we choose α and β such that 4 < α < β < 7, then when x > 7:
- (x - 4) > 0
- (x - 7) > 0
- (x - α) > 0
- (x - β) > 0
- The product of these factors is positive, which does not satisfy the condition for x > 7.
Therefore, our quartic polynomial inequality (x - 4)(x - 7)(x - α)(x - β) < 0 matches the solution x < 4 and x > 7.