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Chapter 3: Problem 25

For each function: a. Make a sign diagram for the first derivative. b. Make a sign diagram for the second derivative. c. Sketch the graph by hand, showing all relative extreme points and inflection points. $$ f(x)=\sqrt[4]{x^{3}} $$

Short Answer

Expert verified
The function increases and is concave down for \( x > 0 \) with no inflection points in this range.

Step by step solution

01

Find the First Derivative

First, we need to find the derivative of the given function \( f(x) = \sqrt[4]{x^3} = x^{3/4} \). The derivative is \( f'(x) = \frac{3}{4}x^{-1/4} = \frac{3}{4\sqrt[4]{x}} \) for \( x > 0 \) and undefined for \( x \leq 0 \).
02

Determine the Sign of the First Derivative

Analyze the sign of the first derivative. The expression \( f'(x) = \frac{3}{4\sqrt[4]{x}} \) indicates that for \( x > 0 \), \( f'(x) > 0 \). Therefore, the first derivative is positive for all \( x > 0 \), showing that the function is increasing on this interval.
03

Find the Second Derivative

Now we find the second derivative. First, simplify \( f'(x) = 3/4 \, x^{-1/4} \). The derivative of this gives \( f''(x) = -\frac{3}{16}x^{-5/4} = -\frac{3}{16\sqrt[4]{x^5}} \), which is defined for \( x > 0 \).
04

Determine the Sign of the Second Derivative

Consider the expression \( f''(x) = -\frac{3}{16\sqrt[4]{x^5}} \). Notice that for \( x > 0 \), \( f''(x) < 0 \), showing that the function is concave down on \( x > 0 \).
05

Identify Critical and Inflection Points

Since \( f'(x) \) is undefined at \( x = 0 \), it is a potential critical point. There are no changes in concavity since \( f''(x) < 0 \) for all \( x > 0 \), indicating no inflection points in \( x > 0 \). At \( x = 0 \), the derivative is undefined, impacting only the behavior at the boundary.
06

Sketch the Graph

To sketch the graph: for \( x > 0 \), the function \( f(x) \) is increasing and concave down. It passes through the origin, starting at \( x = 0 \). The curve rises as \( x \) increases, but without points of horizontal tangency or inflection in positive \( x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Derivative
In calculus, the first derivative of a function provides critical information about the function's behavior. When you compute the derivative of a function, you are essentially determining the rate at which the function's value changes as the input changes. The first derivative of our example function, \(f(x) = \sqrt[4]{x^3} = x^{3/4}\), is given by \(f'(x) = \frac{3}{4\sqrt[4]{x}}\) when \(x > 0\). It is undefined for \(x \leq 0\). Understanding the sign of this derivative:
  • If \(f'(x) > 0\), the function is increasing.
  • If \(f'(x) < 0\), the function is decreasing.
In this problem, since \(f'(x) > 0\) for \(x > 0\), the function is consistently increasing in this region. This means as \(x\) increases, the value of \(f(x)\) rises.
Second Derivative
The second derivative provides insights into the concavity of a function. Concavity tells us how the slope of a function's graph changes:
  • If the second derivative \(f''(x) > 0\), the function is concave up (U-shaped).
  • If \(f''(x) < 0\), the function is concave down (n-shaped).
For the function \(f(x) = \sqrt[4]{x^3}\), the second derivative is \(f''(x) = -\frac{3}{16\sqrt[4]{x^5}}\), which is only defined when \(x > 0\). This expression is negative for all \(x > 0\), indicating that the function is concave down in this interval. This means that as \(x\) increases, the function continues to rise but at a decreasing rate.
Critical Points
Critical points are where the function's derivative is zero or undefined. At these points, the function may reach a maximum, minimum, or an inflection point. For \(f(x) = \sqrt[4]{x^3}\):
  • \(f'(x)\) is undefined at \(x = 0\).
  • There are no points where \(f'(x) = 0\) for \(x > 0\), meaning no horizontal tangents exist in this region.
The undefined derivative at \(x = 0\) suggests that we consider the boundary's behavior. Here, \(f(x)\) passes through the origin and starts increasing immediately as \(x\) becomes positive.
Inflection Points
Inflection points occur where the second derivative changes sign, indicating a change in the concavity of the function. A function is most interesting at these points because they signify where the function changes from being concave upwards to concave downwards, or vice versa. However, for \(f(x) = \sqrt[4]{x^3}\), the second derivative \(f''(x) = -\frac{3}{16\sqrt[4]{x^5}}\) is negative for all \(x > 0\). This means there is no sign change and consequently, there are no inflection points for this function when \(x > 0\). The function smoothly increases while maintaining a concave down arc throughout this interval.

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