let f(x) = cube root of x, solve each of the ff equations for x

(f(x))^-1 = 11, then x =
f^-1(x) = 11, then x =
f(x^-1) = 11, then x =

Let's solve each equation step by step, given \( f(x) = \sqrt[3]{x} \).

### 1. \( (f(x))^{-1} = 11 \)

We know that \( f(x) = \sqrt[3]{x} \), so:

\[
(f(x))^{-1} = \frac{1}{f(x)} = 11
\]

This implies:

\[
\frac{1}{\sqrt[3]{x}} = 11
\]

Now, solve for \( x \):

\[
\sqrt[3]{x} = \frac{1}{11}
\]

Cube both sides:

\[
x = \left(\frac{1}{11}\right)^3 = \frac{1}{11^3} = \frac{1}{1331}
\]

Thus, \( x = \frac{1}{1331} \).

---

### 2. \( f^{-1}(x) = 11 \)

\( f^{-1}(x) \) represents the inverse of \( f(x) \), where \( f(x) = \sqrt[3]{x} \). The inverse of \( f(x) \) is \( f^{-1}(x) = x^3 \).

Given:

\[
f^{-1}(x) = 11
\]

This means:

\[
x^3 = 11
\]

Solve for \( x \):

\[
x = \sqrt[3]{11}
\]

So, \( x = \sqrt[3]{11} \) (or approximately \( x \approx 2.223 \)).

---

### 3. \( f(x^{-1}) = 11 \)

We know \( f(x) = \sqrt[3]{x} \), so:

\[
f(x^{-1}) = \sqrt[3]{x^{-1}} = \frac{1}{\sqrt[3]{x}}
\]

Given:

\[
\frac{1}{\sqrt[3]{x}} = 11
\]

This is the same as the first equation:

\[
\sqrt[3]{x} = \frac{1}{11}
\]

Cube both sides:

\[
x = \frac{1}{11^3} = \frac{1}{1331}
\]

Thus, \( x = \frac{1}{1331} \).

---

### Summary of Solutions:
1. \( (f(x))^{-1} = 11 \Rightarrow x = \frac{1}{1331} \)
2. \( f^{-1}(x) = 11 \Rightarrow x = \sqrt[3]{11} \)
3. \( f(x^{-1}) = 11 \Rightarrow x = \frac{1}{1331} \)

Would you like more details or explanations?

Here are 5 related questions to expand on this:

1. How do you calculate the inverse of a function?
2. What are the properties of cube roots in solving equations?
3. How do you solve equations with exponents in the denominator?
4. What is the significance of a function’s inverse in algebra?
5. How do you interpret and solve \( f(x^{-1}) \)?

**Tip:** Always check whether you need to take the inverse of the function or the inverse of the variable before solving.

let f(x) = cube root of x, solve each of the ff equations for x
(f(x))^-1 = 11, then x =
f^-1(x) = 11, then x =
f(x^-1) = 11, then x =

Learn how to solve equations involving cube roots and inverse functions. This problem explores algebraic concepts such as cube roots, inverse functions, and solving equations step-by-step. Suitable for students in Grades 10-12.

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