Problem 6 - Calculus with Concepts in Calculus

Question

Find the intervals on which the graph of the function is concave upward and those on which it is concave downward. $$ g(x)=x \sqrt{x-1} $$

Step-by-Step Solution

Answer

The function is concave down on \((1, \frac{4}{3})\) and concave up on \((\frac{4}{3}, \infty)\).

Step 1: Find the First Derivative

To find where the function is concave up or down, we first need the first derivative. Let's find \( g'(x) \):Firstly, rewrite \( g(x) \) as \( x(x-1)^{1/2} \). Use the product rule for derivatives, which states \( (uv)' = u'v + uv' \), where \( u = x \) and \( v = (x-1)^{1/2} \):- \( u' = 1 \)- \( v' = \frac{1}{2}(x-1)^{-1/2} \)Substitute these derivatives into the product rule: \[ g'(x) = 1 \cdot (x-1)^{1/2} + x \cdot \frac{1}{2}(x-1)^{-1/2} \]Simplify the expression to get:\[ g'(x) = (x-1)^{1/2} + \frac{x}{2\sqrt{x-1}} \]

Step 2: Simplify the First Derivative

Let's make \( g'(x) \) easier to work with:Notice \( (x-1)^{1/2} = \sqrt{x-1} \), so we rewrite:\[ g'(x) = \sqrt{x-1} + \frac{x}{2\sqrt{x-1}} \]Bring these terms to a common denominator: \[ g'(x) = \frac{2(x-1) + x}{2\sqrt{x-1}} = \frac{3x - 2}{2\sqrt{x-1}} \]

Step 3: Find the Second Derivative

To determine concavity, we need the second derivative \( g''(x) \). Differentiate \( g'(x) = \frac{3x - 2}{2\sqrt{x-1}} \) using the quotient rule:The quotient rule is \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \), where \( u = 3x - 2 \) and \( v = 2\sqrt{x-1} \):- \( u' = 3 \)- \( v = 2(x-1)^{1/2}, v' = \frac{2}{2}(x-1)^{-1/2} = \frac{1}{\sqrt{x-1}} \)Apply the quotient rule:\[ g''(x) = \frac{3 \cdot 2\sqrt{x-1} - (3x-2)\cdot\frac{1}{\sqrt{x-1}}}{(2\sqrt{x-1})^2} \]Simplify:\[ g''(x) = \frac{6\sqrt{x-1} - \frac{3x-2}{\sqrt{x-1}}}{4(x-1)} = \frac{6(x-1) - (3x-2)}{4(x-1)\sqrt{x-1}} \]Combine terms:\[ g''(x) = \frac{6x - 6 - 3x + 2}{4(x-1)\sqrt{x-1}} = \frac{3x - 4}{4(x-1)\sqrt{x-1}} \]

Step 4: Determine Concavity Intervals

We need to find where \( g''(x) \) changes sign to determine concavity:1. Set numerator \( 3x - 4 = 0 \) to find critical points: \[ 3x - 4 = 0 \] Solves to \( x = \frac{4}{3} \)2. Examine the intervals \((1, \frac{4}{3})\) and \((\frac{4}{3}, \infty)\) since function is undefined for \(x \leq 1\): - Choose a test point in \((1, \frac{4}{3})\), e.g., \(x = 1.25\), and evaluate \(g''(x) < 0\) indicating concave down. - Choose a test point in \((\frac{4}{3}, \infty)\), e.g., \(x = 2\), and evaluate \(g''(x) > 0\) indicating concave up.

Key Concepts

First Derivative

When we talk about the first derivative in calculus, we mean the expression that gives us the slope or rate of change of a function at any given point. It's like finding out how steep or flat a road is at a specific location. The first derivative tells us about the direction the graph of the function is heading.
To find the first derivative of a function such as \( g(x) = x \sqrt{x-1} \), we often employ rules like the product rule if the function is composed of multiple parts being multiplied. Remember, the product rule says that if you have two functions, say \( u \) and \( v \), then the derivative of their product is \( (uv)' = u'v + uv' \).
In our scenario where \( u = x \) and \( v = (x-1)^{1/2} \), you calculate as follows:

• Differentiate \( u \): \( u' = 1 \)
• Differentiate \( v \): \( v' = \frac{1}{2}(x-1)^{-1/2} \)

Substitute these into the product rule formula to get the first derivative:\[ g'(x) = (x-1)^{1/2} + \frac{x}{2\sqrt{x-1}} \]This tells us how the function behaves locally.

Second Derivative

The second derivative provides deeper insights into a function's behavior. While the first derivative gave us the slope, the second derivative tells us about the concavity, or the bending of the function's graph. Essentially, it indicates whether our function is curving up or down at any point.
To find the second derivative of \( g(x) = x \sqrt{x-1} \), we begin by differentiating the first derivative using additional calculus rules. In this exercise, applying the quotient rule becomes essential.
The quotient rule formula is applied as follows:\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]Here, \( u = 3x - 2 \) and \( v = 2\sqrt{x-1} \). Calculating their derivatives:

• \( u' = 3 \)
• \( v' = \frac{1}{\sqrt{x-1}} \)

Then, substitute into the quotient rule to find \( g''(x) \):\[ g''(x) = \frac{3 \cdot 2\sqrt{x-1} - (3x - 2)\cdot \frac{1}{\sqrt{x-1}}}{(2\sqrt{x-1})^2} \]Simplifying gives:\[ g''(x) = \frac{3x - 4}{4(x-1)\sqrt{x-1}} \]This expression is critical for checking where the function is concave.

Quotient Rule

The quotient rule is a powerful tool in calculus used when differentiating functions that are a division of two functions. Like a recipe, it requires precise steps to combine the ingredients correctly! It’s especially useful for functions like \( g(x) = \frac{3x-2}{2\sqrt{x-1}} \).
When performing the quotient rule, take note of the formula: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]This lets you find the derivative of \( \frac{u}{v} \) by using:

• The derivative of the numerator \( u \), which is \( u' \)
• The derivative of the denominator \( v \), which is \( v' \)

In this exercise, our numerator \( u = 3x - 2 \) and our denominator \( v = 2\sqrt{x-1} \), hence:

• \( u' = 3 \)
• \( v' = \frac{1}{\sqrt{x-1}} \)

Applying the quotient rule, substitution gives:\[ g''(x) = \frac{3 \cdot 2\sqrt{x-1} - (3x-2)\cdot \frac{1}{\sqrt{x-1}}}{4(x-1)} \]This technique ensures we handle divisions in derivations without mistakes.

Product Rule

The product rule is essential for finding derivatives of functions that are products of two sub-functions. If ever faced with multiplying functions like \( g(x) = x \cdot (x-1)^{1/2} \), this rule is your go-to.
It’s expressed simply as:\[ (uv)' = u'v + uv' \]Here, you differentiate each function component separately:

• For \( u = x \), you find \( u' = 1 \)
• For \( v = (x-1)^{1/2} \), you calculate \( v' = \frac{1}{2}(x-1)^{-1/2} \)

As you combine these, you get:\[ g'(x) = 1 \cdot (x-1)^{1/2} + x \cdot \frac{1}{2}(x-1)^{-1/2} \]Thus giving:\[ g'(x) = \sqrt{x-1} + \frac{x}{2\sqrt{x-1}} \]This approach changes the way complex products transform into simpler expressions for analysis.