Determine the value of k so that the function f(x)=(x-6)/(2x+k) will be its own inverse?

y = (x - 6) / (2x + k)

x = (y - 6) / (2y + k)

x * (2y + k) = y - 6

2xy + xk = y - 6

xk + 6 = y - 2xy

xk + 6 = y * (1 - 2x)

(xk + 6) / (1 - 2x) = y

(xk + 6) / (1 - 2x) = (x - 6) / (2x + k)

(xk + 6) * (2x + k) = (x - 6) * (1 - 2x)

2kx^2 + 12x + k^2 * x + 6k = x - 2x^2 - 6 + 12x

2kx^2 + (12 + k^2) * x + 6k = -2x^2 + 13x - 6

2k = -2

k = -1

12 + k^2 = 13

k^2 = 1

k = -1 , 1

6k = -6

k = -1

k = -1 seems to work.

The easiest way is to just look at the asymptotes.

f(x)=(x-6)/(2x+k)  has the horizontal asymptote y = 1/2.  For the function to be invertible it must then have x = 1/2 as a vertical asymptote.

So you want 2x+k to equal 0 when x is 1/2.

2*1/2 + k = 0

k = -1

f(g) = x = (g-6)/(2g+k)

2gx + kx = g-6

2gx - g = -6-kx

g(2x-1) = -(6+kx)

g = -(6+kx)/(2x-1)

g is the inverse of f, so if f is to be its own inverse, then f = g

f(x) = (x-6)/(2x+k) = -(6+kx)/(2x-1) = g(x)

(x-6)(2x-1) = -(6+kx)(2x+k)

2x² - x - 12x + 6 = -(12x+6k+2kx²+k²x)

2x² - 13x + 6 = -[2kx² + (12+k²)x + 6k)

2x² + 2kx² - 13x + (12+k²)x + 6+6k = 0

2(k+1)x² + (k²-1)x + 6(k+1) = 0

(k+1)[2x² + (k-1)x + 6] = 0

k+1 = 0

k = -1

Ans: k = -1

f(x)=(x-6)/(2x+k)

f(x) is its own inverse.

f(x)^2=1=(x-6)^2/(2x+k)^2, (2x+k)^2=(x-6)^2

(2x+k)^2-(x-6)^2=0

(2x+k) - (x-6) =0, (2x+k)+ (x-6)=0

x+k+6=0 , k= -(x+6)

The second equation gives

3x+k-6=0, k=6-3x.

These are two values for which the inverse can be seen.

Don't you substitute f(x) = y and change the y = x and then simply and compare the two equations?

You have y = (x-6)/(2x+k).

Solving for x you get

x = (-ky-6)/(2y-1) right? Confirm this and we will proceed. Else we assume you now have figured it out & you're done!