The question asks you to solve parts a and b for the given one-to-one function $f(x) = x^3 - 6$. Let's work through these step by step.

a. Find $f(1)$

The function is $f(x) = x^3 - 6$. Substituting $x = 1$: $f(1) = 1^3 - 6 = 1 - 6 = -5$ So, $f(1) = -5$.

b. Find $f^{-1}(7)$

The inverse $f^{-1}(7)$ means finding the value of $x$ such that $f(x) = 7$. Start with: $f(x) = x^3 - 6 = 7$ Add 6 to both sides: $x^3 = 13$ Now take the cube root of both sides: $x = \sqrt[3]{13}$ Thus, $f^{-1}(7) = \sqrt[3]{13}$ or approximately $2.35$ as a decimal.

Final Answers:

• a. $f(1) = -5$
• b. $f^{-1}(7) = \sqrt[3]{13}$ or $\approx 2.35$.

Would you like me to explain any steps in more detail or provide further clarification?

Related Questions:

1. How do you determine if a function has an inverse?
2. What properties make a function one-to-one?
3. Can the inverse function $f^{-1}(x)$ be directly calculated without solving for $x$?
4. How would you graph both $f(x)$ and $f^{-1}(x)$?
5. What happens if the function is not one-to-one? How does this affect the inverse?

Tip:

When working with inverses, ensure that the original function is one-to-one by checking that it passes the Horizontal Line Test. This guarantees the existence of an inverse.