1. Some functions that aren't invertible can be made invertible by restricting their domains. For example, the function \(x^2\) is invertible if we restrict \(x\) to the interval \([0,\infty)\), or to any subset of that interval. In that case, the inverse function is \(\sqrt x\). (We could also restrict \(x^2\) to the domain \((-\infty,0]\), in which case the inverse function would be \(-\sqrt{x}\).)
Similarly, by restricting the domain of the function \(f(x) = 2x^2-4x-5\) to an interval, we can make it invertible. What is the largest such interval that includes the point \(x=0\)?
2. The function \(f(x) = \frac{cx}{2x+3}\)
satisfies \(f(f(x))=x\), \(x\ne -\frac 32\) for all real numbers . Find \(c\).
+128407
1. The vertex of this parabola is (1, -7)....so.....restricting the domain to (-infinity, 1 ] will make the function invertible...... and x = 0 lies within this interval
2. If f ( f(x) ) = x we can write
c ( (cx) / [2x + 3] )
________________ = x multiply throgh by the denominator on the left side
2 ( (cx/ [2x + 3]) + 3
c( (cx)) / [2x + 3] ) = x [ 2 ( (cx) / [2x + 3] ) + 3] simplify
c^2x / [2x + 3] = x [ 2cx + 6x + 9 ] / [2x + 3]
c^2x = 2cx^2 + 6x^2 + 9x
2cx^2 + 6x^2 + (9 - c^2) x = 0
Note....that for any x, this will equal 0 whenever
2c + 6 + (9-c^2) = 0
-c^2 + 2c + 15 = 0 multiply through by -1
c^2 - 2c - 15 = 0 factor
(c - 5) (c + 3) = 0
So....setting each factor to 0 and solving for c we have the possible values c = 5 or c = -3
Test c = 5 in f(f(x))
5[5x /[2x + 3] ] [ 25x ] / [2x + 3]
_______________ = _________________ = 25x / [ 16x + 9 ]
2 [ 5x / [2x + 3]] + 3 [10x + 6x + 9]/ [2x + 3]
So c = 5 is not a solution
Test c = -3 in f (f (x))
-3 [ -3x / [2x + 3] ] [ 9x] / [2x + 3] 9x
________________ = __________________ = ___ = x
2 [ -3x / [2x + 3] ] + 3 [-6x + 6x + 9] / [2x + 3] 9
So.... c = -3
