To find [tex]\( x \)[/tex] for which [tex]\( f^{-1}(x) = 1 \)[/tex], follow these steps:
1. Understand the Function Relation:
Since [tex]\( f^{-1}(x) = 1 \)[/tex], it means that [tex]\( f(1) = x \)[/tex]. Therefore, we need to solve the equation [tex]\( 7x^3 + 2x + 2 = 1 \)[/tex].
2. Set up the Equation:
[tex]\[ 7x^3 + 2x + 2 = 1 \][/tex]
3. Simplify the Equation:
Subtract 1 from both sides to set the equation to 0:
[tex]\[ 7x^3 + 2x + 1 = 0 \][/tex]
4. Solving the Cubic Equation:
To find the roots of the cubic equation [tex]\( 7x^3 + 2x + 1 = 0 \)[/tex], we can use various algebraic or numerical methods. Here, we directly present the complex roots found:
[tex]\[
x = \left[ \frac{2}{7 \cdot \left(-\frac{1}{2} - \frac{\sqrt{3}i}{2}\right) \left(\frac{27}{14} + \frac{3\sqrt{4641}}{98}\right)^{\frac{1}{3}}} - \frac{\left(-\frac{1}{2} - \frac{\sqrt{3}i}{2}\right) \left(\frac{27}{14} + \frac{3\sqrt{4641}}{98}\right)^{\frac{1}{3}}}{3}, \quad -\frac{\left(-\frac{1}{2} + \frac{\sqrt{3}i}{2}\right) \left(\frac{27}{14} + \frac{3\sqrt{4641}}{98}\right)^{\frac{1}{3}}}{3} + \frac{2}{7 \cdot \left(-\frac{1}{2} + \frac{\sqrt{3}i}{2}\right) \left(\frac{27}{14} + \frac{3\sqrt{4641}}{98}\right)^{\frac{1}{3}}}, \quad -\frac{\left(\frac{27}{14} + \frac{3\sqrt{4641}}{98}\right)^{\frac{1}{3}}}{3} + \frac{2}{7 \left(\frac{27}{14} + \frac{3\sqrt{4641}}{98}\right)^{\frac{1}{3}}} \right]
\][/tex]
Thus, the solutions for [tex]\( x \)[/tex] which satisfy the equation [tex]\( f^{-1}(x) = 1 \)[/tex] are the three roots of the cubic equation given above. Each root involves a combination of constants and complex numbers, showing that the solutions are non-trivial and involve complex arithmetic.
