Answer :
The values of x such that function f(x) is "above" function g(x) is represented by following inequalities: - 10 < x < 0 and x > 11.
How to find the values of x such that a function is above another one
In this question we see the representation of two functions, of which we must determine the values of x such that function f(x) is "above" function g(x). Mathematically speaking, f(x) is a linear function and g(x) is a polynomic function. Given the graph of the two functions, we can estimate the values of x such that f(x) > g(x), that is, in the intervals such that f(x) is "above".
This requirement is fulfilled when x is greater than - 10 and less than 0 and when x is greater than 11. This is represented by following inequalities:
- 10 < x < 0 and x > 11.
The values of x for which f(x) > g(x) are -1 < x < 2/3.
To find the points of intersection, we set f(x) equal to g(x) and solve for x:
3x² + 4x + 6 = 3x + 4
Subtract 3x + 4 from both sides:
3x² + 4x + 6 - (3x + 4) = 0
This simplifies to:
3x² + 4x - 3x - 2 = 0
Now, combine like terms:
3x² + x - 2 = 0
Next, we can factor the quadratic equation:
(3x - 2)(x + 1) = 0
Now, we solve for x:
3x - 2 = 0
3x = 2
x = 2/3
x + 1 = 0
x = -1
So, we have two points of intersection at x = 2/3 and x = -1.
Now, we need to determine the regions where f(x) > g(x). We can do this by checking the sign of the difference between f(x) and g(x) in three intervals:
Interval 1: x < -1
Interval 2: -1 < x < 2/3
Interval 3: x > 2/3
In Interval 1, since x < -1, both f(x) and g(x) are negative, but f(x) is greater because it's a quadratic function.
In Interval 2, -1 < x < 2/3, g(x) is negative, and f(x) is positive. Therefore, f(x) > g(x) in this interval.
In Interval 3, x > 2/3, both f(x) and g(x) are positive, but f(x) is greater due to its quadratic nature.
So, the values of x for which f(x) > g(x) are: -1 < x < 2/3.
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