Let's go through the problem step by step, starting with the relevant functions:

• $f(x) = \frac{1}{x} - 3$
• $g(x) = x^3$

Solving Exercises 83-88

83. $(f^{-1} \circ g^{-1})(1)$
This means we need to find $f^{-1}(g^{-1}(1))$. We'll first find the inverse functions of $f(x)$ and $g(x)$, then evaluate the composition.

1. Find the inverse of $g(x) = x^3$: $g^{-1}(x) = \sqrt[3]{x}$

2. Find the inverse of $f(x) = \frac{1}{x} - 3$: Set $y = \frac{1}{x} - 3$, then solve for $x$: $y + 3 = \frac{1}{x}$ $x = \frac{1}{y + 3}$ So, $f^{-1}(x) = \frac{1}{x + 3}$.

3. Now, compute $g^{-1}(1) = \sqrt[3]{1} = 1$.

4. Plug this into $f^{-1}$: $f^{-1}(1) = \frac{1}{1 + 3} = \frac{1}{4}$

Thus, $(f^{-1} \circ g^{-1})(1) = \frac{1}{4}$.

84. $(g^{-1} \circ f^{-1})(-3)$

1. From earlier, we know $f^{-1}(x) = \frac{1}{x + 3}$.
2. Compute $f^{-1}(-3) = \frac{1}{-3 + 3} = \text{undefined}$ (division by 0).

So, $(g^{-1} \circ f^{-1})(-3)$ is undefined.

85. $(f^{-1} \circ f)(6)$
Since $f^{-1}(f(x)) = x$, this simplifies to: $(f^{-1} \circ f)(6) = 6$

86. $(g^{-1} \circ g^{-1})(-4)$
First, compute $g^{-1}(-4) = \sqrt[3]{-4}$. There's no simple cube root of $-4$, so we leave this as $\sqrt[3]{-4}$.

Next, compute $g^{-1}(\sqrt[3]{-4})$, which would be the cube root of $\sqrt[3]{-4}$, or $(-4)^{1/9}$, an irrational number. The exact value is complex, but it's approximately $-1.1487$.

87. $(f \circ g)^{-1}$
This asks for the inverse of the composition. First, find $f(g(x))$: $f(g(x)) = f(x^3) = \frac{1}{x^3} - 3$ To find the inverse of $f(g(x))$, let $y = \frac{1}{x^3} - 3$, then solve for $x$. This would be a more involved process requiring algebraic manipulation.

Moving to Exercises 89-92 (using new functions):

• $f(x) = x + 4$
• $g(x) = 2x - 5$

89. $g^{-1} \circ f^{-1}$
This requires first finding the inverses of both functions and then composing them.

Would you like me to continue solving the remaining exercises in detail or expand on one of these steps?

5 Related Questions:

1. How do you find the inverse of a general function?
2. Why is $f^{-1}(x) = \frac{1}{x+3}$ the inverse of $f(x)$?
3. What happens if a function is not invertible at a certain point (e.g., $x = -3$)?
4. Can the cube root of a negative number be calculated, and how?
5. How does function composition affect the domain and range of the resulting function?

Tip:

When finding the inverse of a composition of functions, try breaking the problem down by solving the innermost function's inverse first and then moving outward.