Problem 65 Use a calculator to determine wh... [FREE SOLUTION] (2024)

Expert verified

The cube root function, as seen in our exercise with the given function \(f(x) = \sqrt[3]{x-5}\), transforms the input by taking its cube root. Unlike square roots, cube roots apply to all real numbers, including negatives. This function is continuous and smooth for all real numbers, meaning it doesn't have sudden jumps or breaks within its domain.
However, when we're specifically dealing with the cube root of a shifted variable, like \(x-5\), we're focusing on translating the entire cube root curve horizontally. This shift is crucial to understanding behaviors and where certain properties, such as derivatives, may not exist. Keep in mind:

• The domain of \(\sqrt[3]{x-5}\) includes all real numbers.
• This function is positive when \(x-5 > 0\), zero when \(x-5 = 0\), and negative when \(x-5 < 0\).
• It smoothly continues without any abrupt changes in direction.

Understanding this fundamental behavior helps in analyzing further characteristics like differentiation.

The chain rule is a fundamental tool in calculus for differentiating composite functions. In our exercise, the given function is a composite where \(u = x-5\), making \(f(x)\) a function of another function (\(u\)).
The chain rule formula is given by:

• If \(h(x) = f(g(x))\), then \(h'(x) = f'(g(x)) \cdot g'(x)\).

Applying the chain rule, we start by identifying the inner and outer functions. Here, \(g(x) = x-5\) (the inner function) and \(f(u) = \sqrt[3]{u}\) (the outer function).
We first find the derivative of the outer function:\(f'(u) = \frac{1}{3} u^{-\frac{2}{3}}\). Now, applying it to our specific \(u = x-5\), we get \(f'(u) = \frac{1}{3} (x-5)^{-\frac{2}{3}}\).
Next, we need the derivative of the inner function, which is simply \(g'(x) = 1\).
Using the chain rule, we multiply these together:

• \(f'(x) = f'(g(x)) \cdot g'(x) = \frac{1}{3}(x-5)^{-\frac{2}{3}} \cdot 1 = \frac{1}{3}(x-5)^{-\frac{2}{3}}\).

This calculated derivative helps us understand the behavior of our original function around certain points like \(x=5\).

In calculus, a derivative at a certain point can be undefined for various reasons. For our function \(f(x) = \sqrt[3]{x-5}\), the derivative \(f'(x) = \frac{1}{3}(x-5)^{-\frac{2}{3}}\) is undefined when the expression inside the parentheses is zero.
This expression, \((x-5)^{-\frac{2}{3}}\), involves raising \(x-5\) to the power of \(-\frac{2}{3}\). When the base (\(x-5\)) is zero, the whole expression becomes undefined. Thus, we solve:

• \(x-5 = 0\)
• \(x = 5\).

Therefore, at \(x = 5\), the derivative is undefined. This is because the function's slope becomes infinitely steep or vertical. The cube root function itself remains continuous and defined, but its rate of change (slope) is not clearly determinable at this exact point.
Understanding undefined derivatives is essential in grasping the full behavior of functions, especially at critical points where sudden changes in behavior occur. It tells us that while the function continues to exist, differentiating it at that specific point is not possible.