(f-g) (fx)=1/x+2 g(x)= 3x
I got -3x^2-6x+1/x+2
is this correct?
I'm assuming x+2 is the denominator:
Given: F(x) = 1 / (x+2); g(x) = 3x.
Find: (f-g)(x).
(F-g)(x) = 1 / (x+2) - 3x; LCD = (x+2).
1 / (x+2) - 3x(x+2) / (x+2)=
(1-3x^2-6x)/(x+2)=
(-3x^2-6x+1)/(x+2). Numerator cannot be
factored. So our answers are identical.
Thanks hon!
You are welcome!
To determine if your result is correct, let's start by rearranging and simplifying the given equation: (f - g) * (fx) = 1/(x + 2).
The first step is to express f - g in terms of x.
Since f(x) = (f - g)(x), we can substitute g(x) = 3x into the expression to obtain f(x) = (f - 3x)(x).
Next, we substitute these values into the given equation: (f - 3x)(x) * fx = 1/(x + 2).
Expanding the left side of the equation, we have (f - 3x)(x)(fx). Note that fx means the function f applied to x, which is equal to f(x).
Multiplying, we get (f^2)x - 3x(fx) = 1/(x + 2).
Now, let's substitute our expression for f(x): (f - 3x)(x)(fx) = (f^2 - 3x^2)x = 1/(x + 2).
Expanding again, we have (f^2)x - 3x^2x = 1/(x + 2).
Simplifying further, we obtain f^2x - 3x^3 = 1/(x + 2).
Finally, we can rearrange this equation and solve for f(x): f^2x = 1/(x + 2) + 3x^3.
Dividing both sides by x gives us f^2 = (1/(x(x + 2))) + 3x^2.
Now, substituting the value of f(x) we found earlier, f(x) = (f - 3x)(x), we have: (f - 3x)(x)^2 = (1/(x(x + 2))) + 3x^2.
Since this is a quadratic equation (f - 3x)(x)^2 = (1/(x(x + 2))) + 3x^2, it is difficult to solve for f(x) algebraically without additional information or constraints.
Therefore, it is not possible to determine if your result, -3x^2 - 6x + (1/x + 2), is correct based on the given equation.