Question

In Exercises \(67-70,\) use the functions \(f(x)=\frac{1}{8} x-3\) and \(g(x)=x^{3}\) to find the indicated value. $$ \left(f^{-1} \circ g^{-1}\right)(1) $$

Short answer

The value of \( \left(f^{-1} \circ g^{-1}\right)(1) \) is 32

Step-by-step solution

Find the Inverse of f(x) and g(x)

The given functions are \(f(x) = \frac{1}{8}x - 3\) and \(g(x) = x^{3}\). We need to solve each function for x to find their inverses. For \(f(x) = \frac{1}{8}x - 3\), multiply both sides by 8 to get \(8f(x) = x - 24\), then add 24 to both sides to get \(f^{-1}(x) = 8x + 24\). For \(g(x) = x^{3}\), the inverse function \(g^{-1}(x)\) is simply \(\sqrt[3]{x}\) or \(x^{1/3}\).

Calculate \(g^{-1}(1)\)

Now we substitute x = 1 into the inverse function for g(x), which is \(g^{-1}(1) = (1)^{1/3} = 1\).

Final Calculation

Finally, substitute the result from Step 2 into \(f^{-1}(x)\), which is \(f^{-1}(1) = 8*1 + 24 = 32\). That is the final answer.

Key concepts

Composite Functions

When we combine two functions by applying one function to the result of another, we create what's called a composite function. This process can be denoted as \( (f \circ g)(x) \), which reads as 'f of g of x'. To evaluate a composite function, you first apply the function g to x to get g(x), and then you take that result and apply the function f to it to get f(g(x)).

In calculus, understanding composite functions is crucial for operations such as function composition and chaining, which lead to the chain rule, a method for finding derivatives of composite functions. When dealing with the inverse of composite functions, as in our exercise, the process is also known as function inversion. You'll first find the inverse of both functions involved and then perform the composite operation in a reversed order.

Function Inversion

The inverse of a function, often denoted as \(f^{-1}(x)\), is a fundamental concept where we reverse the original function's input-output pairs. Simply put, if you have a function that normally takes an input 'a' and gives an output 'b', the inverse function would take 'b' as an input to give 'a' as an output.

The process of finding this inverse involves swapping the dependent and independent variables and then solving for the new independent variable. In our example, to find the inverse of the function \( f(x) = \frac{1}{8}x - 3 \), we swap x and \( f(x) \) and solve for x. Function inversion is essential for solving equations and understanding the symmetry of functions along the line \( y=x \).

Calculus Problems

Solving calculus problems often involves multiple steps that require an understanding of different calculus concepts. For instance, you might need to find the derivative of a function, calculate the area under a curve, or, as with our problem, determine the value of a composite of inverse functions.

Effective strategies include breaking down the problem into smaller steps, making sure to understand the function behavior, and then applying the appropriate rules and operations. Always look out for function inverses, roots, and powers as they indicate typical calculus operations. Practice is key to mastering these problems, along with a solid grasp of the fundamental concepts.

Power Function

A power function is a type of mathematical function of the form \( f(x) = x^n \), where \( n \) is a real number. The inverse of a power function with an odd exponent, like the \( g(x) = x^3 \) in our problem, is another power function \( g^{-1}(x)=x^{\frac{1}{3}} \). This is because the cube root is the inverse operation to cubing a number.

Understanding power functions is essential in calculus because they are often used to model real-world phenomena. Their simplicity makes them useful for demonstrating basic calculus principles such as derivatives and integrals. In our exercise, the cubic function and its inverse play a significant role in the solution process.