Math Problem Statement

Use the Inverse Function Theorem (Theorem 2.3 in the lecture notes) to show that f and g are inverses of each other. f (x) = x^3 + 1 g(x) = cubic root of √(x − 1)

Solution

To show that f(x)=x3+1f(x) = x^3 + 1 and g(x)=x13g(x) = \sqrt[3]{x - 1} are inverses of each other using the Inverse Function Theorem, we'll follow these steps:

1. Check the compositions: f(g(x))f(g(x)) and g(f(x))g(f(x))

For two functions to be inverses, their compositions must satisfy:

f(g(x))=xandg(f(x))=x.f(g(x)) = x \quad \text{and} \quad g(f(x)) = x.

Let's compute each composition:

f(g(x))f(g(x))

We are given g(x)=x13g(x) = \sqrt[3]{x - 1}. Now, let's substitute g(x)g(x) into f(x)=x3+1f(x) = x^3 + 1:

[ f(g(x)) = f\left(\sqrt[3]{x - 1}\right) = \left( \sqrt[3]{x - 1} \right)^3 + 1. ]

Using the property of cube roots, (x13)3=x1\left( \sqrt[3]{x - 1} \right)^3 = x - 1. Thus:

f(g(x))=(x1)+1=x.f(g(x)) = (x - 1) + 1 = x.

g(f(x))g(f(x))

Now, substitute f(x)=x3+1f(x) = x^3 + 1 into g(x)=x13g(x) = \sqrt[3]{x - 1}:

[ g(f(x)) = g(x^3 + 1) = \sqrt[3]{(x^3 + 1) - 1} = \sqrt[3]{x^3}. ]

Since x33=x\sqrt[3]{x^3} = x, we have:

g(f(x))=x.g(f(x)) = x.

2. Apply the Inverse Function Theorem

The Inverse Function Theorem states that if ff is continuously differentiable and its derivative f(x)f'(x) is nonzero at a point, then ff has a local inverse near that point. Moreover, the derivative of the inverse function g(x)g(x) at a point xx can be expressed as:

g(x)=1f(g(x)).g'(x) = \frac{1}{f'(g(x))}.

Check if f(x)f'(x) is nonzero

First, compute f(x)f'(x), the derivative of f(x)=x3+1f(x) = x^3 + 1:

f(x)=3x2.f'(x) = 3x^2.

Since 3x23x^2 is nonzero for all x0x \neq 0, ff is invertible for x0x \neq 0 by the Inverse Function Theorem.

Verify using the Inverse Function Theorem

If ff and gg are inverses, we expect g(x)=1f(g(x))g'(x) = \frac{1}{f'(g(x))}.

  • First, find g(x)g'(x). Since g(x)=x13g(x) = \sqrt[3]{x - 1}, use the chain rule:

g(x)=13(x1)23.g'(x) = \frac{1}{3} \cdot (x - 1)^{-\frac{2}{3}}.

  • Now, check 1f(g(x))\frac{1}{f'(g(x))}:

Since f(x)=3x2f'(x) = 3x^2, substitute g(x)=x13g(x) = \sqrt[3]{x - 1}:

f(g(x))=3(x13)2=3(x1)23.f'(g(x)) = 3 \left( \sqrt[3]{x - 1} \right)^2 = 3 (x - 1)^{\frac{2}{3}}.

Thus:

1f(g(x))=13(x1)23.\frac{1}{f'(g(x))} = \frac{1}{3 (x - 1)^{\frac{2}{3}}}.

This matches g(x)=13(x1)23g'(x) = \frac{1}{3} (x - 1)^{-\frac{2}{3}}, confirming that ff and gg are indeed inverses of each other by the Inverse Function Theorem.

Conclusion

We have shown that f(x)=x3+1f(x) = x^3 + 1 and g(x)=x13g(x) = \sqrt[3]{x - 1} are inverses of each other both by checking their compositions and verifying the conditions of the Inverse Function Theorem.


Would you like further details or have any specific questions on this?

Here are some related questions:

  1. How does the Inverse Function Theorem guarantee the existence of an inverse?
  2. What happens if f(x)=0f'(x) = 0 at a certain point?
  3. Can we apply the Inverse Function Theorem to functions of multiple variables?
  4. What role does continuity play in the Inverse Function Theorem?
  5. How would the analysis change if f(x)f(x) had a more complex form, like f(x)=x4+xf(x) = x^4 + x?

Tip: When using the Inverse Function Theorem, always ensure that the derivative of the function is nonzero at the points of interest.

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Math Problem Analysis

Mathematical Concepts

Inverse Functions
Derivatives
Compositions of Functions

Formulas

f(x) = x^3 + 1
g(x) = \sqrt[3]{x - 1}
f'(x) = 3x^2
g'(x) = \frac{1}{3} (x - 1)^{-\frac{2}{3}}
g'(x) = \frac{1}{f'(g(x))}

Theorems

Inverse Function Theorem

Suitable Grade Level

Undergraduate (Calculus II or III)