Problem 39 - Calculus Single Variable

Question

Find each critical point \(c\) of the given function \(f\). Then use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=(x-4)^{1 / 5}(3 x+1)^{2 / 3} $$

Step-by-Step Solution

Answer

Critical point is at \(x = \frac{37}{19}\) with a local maximum.

Step 1: Find the First Derivative

Start by finding the first derivative of the function \(f(x)=(x-4)^{1/5}(3x+1)^{2/3}\). Use the product rule since it consists of two differentiable functions multiplied together, and chain rule to differentiate nested functions. Let \(u = (x-4)^{1/5}\) and \(v = (3x+1)^{2/3}\). The derivatives are \(u' = \frac{1}{5}(x-4)^{-4/5}\) and \(v' = \frac{2}{9}(3x+1)^{-1/3}\). Applying the product rule, \(f'(x) = u'v + uv'\). Substitute the derivatives and expressions for \(u\) and \(v\) to find \[ f'(x) = \frac{1}{5}(x-4)^{-4/5}(3x+1)^{2/3} + (x-4)^{1/5}\frac{2}{9}(3x+1)^{-1/3}(3). \]

Step 2: Simplify the Derivative

Simplify the expression for \(f'(x)\). Factor out common terms: \((x-4)^{-4/5}\) and \((3x+1)^{-1/3}\). The simplified form becomes:\[ f'(x) = (x-4)^{-4/5}(3x+1)^{-1/3}\left[ \frac{1}{5}(3x+1) + \frac{2}{3}(x-4) \right]. \]Further simplification inside the brackets gives:\[ \frac{1}{5}(3x+1) + \frac{2}{3}(x-4) = \frac{3x + 1}{5} + \frac{2x - 8}{3}. \]Combining fractions:\[ \frac{9x + 3 + 10x - 40}{15} = \frac{19x - 37}{15}. \]

Step 3: Find Critical Points

Set the numerator of the simplified derivative \(f'(x)\) to zero to find critical points, ignoring the denominator since it does not equal zero:\[ 19x - 37 = 0. \]Solve for \(x\):\[ x = \frac{37}{19}. \]

Step 4: Determine the Test Intervals

Create test intervals around the critical point \(x = \frac{37}{19}\), considering the points where the components of \(f'(x)\) are undefined, such as \(x = 4\) and \(x = -\frac{1}{3}\). Define intervals as \((-\infty, -\frac{1}{3})\), \((-\frac{1}{3}, 4)\), and \((4, \infty)\). Use test points from each interval to determine the sign of \(f'(x)\).

Step 5: Apply First Derivative Test

Evaluate \(f'(x)\) at test points like \(x = -1, 0, 5\) for each interval:- For \((-\infty, -\frac{1}{3})\), choose \(x = -1\): \(f'(-1) > 0\).- For \((-\frac{1}{3}, 4)\), choose \(x = 0\): \(f'(0) < 0\).- For \((4, \infty)\), choose \(x = 5\): \(f'(5) > 0\).The function changes from increasing to decreasing at \(x = \frac{37}{19}\), indicating a local maximum.

Key Concepts

Product Rule

The product rule is an essential tool in calculus for differentiating expressions where two functions are multiplied together. Suppose we have two functions, say, \( u(x) \) and \( v(x) \). The product rule states that the derivative of their product \( u(x) \cdot v(x) \) is given by:

• \( (u \cdot v)' = u' v + u v' \)

This formula helps us handle functions like \((x-4)^{1/5}(3x+1)^{2/3}\), where both terms are functions of \(x\). First, differentiate each function separately to find \( u' \) and \( v' \), then apply the product rule to combine them.
In our exercise, for \( u = (x-4)^{1/5} \), we find \( u' = \frac{1}{5}(x-4)^{-4/5} \). For \( v = (3x+1)^{2/3} \), \( v' = \frac{2}{9}(3x+1)^{-1/3} \). Plugging these into the product rule formula allows us to differentiate the original function.

Chain Rule

The chain rule is your go-to guide when dealing with composed or nested functions. If you have a function inside another function, the chain rule helps you differentiate the full composition.
Imagine a function \( g(x) = h(f(x)) \). According to the chain rule, the derivative \( g'(x) \) is given by:

• \( g'(x) = h'(f(x)) \cdot f'(x) \)

This rule is based on the idea of multiplying the derivative of the outer function by the derivative of the inner function.
In our exercise, when we took the derivative of \( u = (x-4)^{1/5} \), we used the chain rule to get \( u' = \frac{1}{5}(x-4)^{-4/5} \). Similarly, for \( v = (3x+1)^{2/3} \), the chain rule helped us determine \( v' = \frac{2}{9}(3x+1)^{-1/3} \). Ensuring accuracy with these steps makes calculating derivatives manageable and reliable.

First Derivative Test

The first derivative test is a method used to determine whether a critical point of a function is a local maximum, a local minimum, or neither. Once you find the critical points by setting the first derivative equal to zero, this test helps verify the behavior of the function around those points.
Here's how it works:

• Choose test points in the intervals around the critical points.
• Evaluate the sign of the derivative \( f'(x) \) at these points.
• Observe how \( f'(x) \) changes from positive to negative or vice versa.

If \( f'(x) \) changes from positive to negative at a critical point, the function has a local maximum there. Conversely, if it changes from negative to positive, there's a local minimum. In our case, at \( x = \frac{37}{19} \), we identified a local maximum because the function shifted from increasing to decreasing.

Simplifying Derivatives

Simplifying derivatives is a crucial step in calculus that makes further calculations much easier. After finding the derivative using the product and chain rules, simplifying can reveal critical points more clearly.
In our exercise, we started with:

• \( f'(x) = \frac{1}{5}(x-4)^{-4/5}(3x+1)^{2/3} + (x-4)^{1/5}\frac{2}{9}(3x+1)^{-1/3}(3) \)

We then factored common terms to simplify:

• \( (x-4)^{-4/5}(3x+1)^{-1/3} \)

Simplifying inside the brackets gave us \( \frac{19x - 37}{15} \), which directly leads to finding critical points by setting this numerator to zero. This simplification made it straightforward to find \( x = \frac{37}{19} \). Reducing complexity in derivatives helps pinpoint important aspects like critical points efficiently.