Let f(x) = x3 − 6x2 + 15x + 3. Then,

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f(x) is invertible f(x) =x3 − 6x2 + 15x + 3 f'(x) = 3 x^2 - 12x + 15 = 3\left( x^2 - 4x + 5 \right) = 3\left( x^2 - 4x + 4 + 1 \right) = 3 \left( x - 2 \right)^2 + $\frac{1}{3}$ > 0 \text { Therefore, f(x) is strictly increasing function }. \Rightarrow f^{- 1} (x) \text { exists } . \text { Hence, f(x) is an invertible function } .

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