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Answer the following.
If f(x) = [4 - (x - 7)³]^(1/5) is a real invertible function, then find f⁻¹(x).
If $f(x) = [4 - (x - 7)^3]^{\frac{1}{5}}$ is a real invertible function, then find $f^{-1}(x)$.
Answer
Answer
Let $y = [4 - (x-7)^3]^{1/5}$.
$y^5 = 4 - (x-7)^3$
$(x-7)^3 = 4 - y^5$
$x = 7 + (4 - y^5)^{1/3}$
Interchanging $x$ and $y$: $f^{-1}(x) = 7 + (4 - x^5)^{1/3}$.
Final answer: $f^{-1}(x) = 7 + (4 - x^5)^{1/3}$