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Sagot :
f(x) as given has no inverse. We see that, for instance, f(1) = f(7) = 32; that is, two different values of x give the same value of f(x), so f is not one-to-one.
We can however restrict the domain over which f is defined to extract an invertible function. One such choice would be to restrict to the interval x ≥ 4, which I'll demonstrate below.
Recall the definition of function inverse:
f(f^(-1)(x)) = x
Then for this function, we have
3 (f^(-1)(x) - 4)² + 5 = x
3 (f^(-1)(x) - 4)² = x - 5
(f^(-1)(x) - 4)² = (x - 5)/3
√((f^(-1)(x) - 4)²)= √((x - 5)/3)
For all x, √(x²) = |x|. Then √((f^(-1)(x) - 4)²) = |f^(-1)(x) - 4|, but by restricting x ≥ 4, or x - 4 ≥ 0, we have the condition that f^(-1)(x) - 4 ≥ 0, and so by definition (of the absolute value function) the absolute value reduces to the positive case. In short, we end up with
f^(-1)(x) - 4 = √((x - 5)/3)
and hence the inverse
f^(-1)(x) = √((x - 5)/3) + 4
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