Let f(x)=3x^2-4x. Find the constant k such that f(x)=f(k-x) for all real numbers x. Thank you!
To find the constant k, we need to set up the equation f(x) = f(k-x) and solve for k.
Given f(x) = 3x^2 - 4x, let's substitute k-x for x in the equation:
f(k-x) = 3(k-x)^2 - 4(k-x)
Expanding and simplifying:
f(k-x) = 3(k^2 - 2kx + x^2) - 4k + 4x
Next, we equate f(x) and f(k-x):
3x^2 - 4x = 3(k^2 - 2kx + x^2) - 4k + 4x
Now, let's simplify both sides of the equation:
3x^2 - 4x = 3k^2 - 6kx + 3x^2 - 4k + 4x
The x^2 terms cancel out, and the x terms will be grouped together:
-4x + 4x = 3k^2 - 6kx - 4k
0 = 3k^2 - 6kx - 4k
Now, let's factor out a k from the right side:
0 = k(3k - 6x - 4)
Since this equation must hold for all real numbers x, the coefficient of the k term must be zero:
3k - 6x - 4 = 0
Simplifying the equation:
3k = 6x + 4
Dividing both sides by 3:
k = (6x + 4) / 3
Therefore, k = 2x + 4/3.
Thus, the constant k is k = 2x + 4/3.