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Chapter 5: Problem 24

In Problems , find c such that \(f^{\prime}(c)=0\) and determine whether \(f(x)\) has a local extremum at \(x=c .\) $$ f(x)=e^{x^{3}} $$

Short Answer

Expert verified
The value of \( c \) is 0, and there is no local extremum at \( x = 0 \).

Step by step solution

01

Find the Derivative of f(x)

To find the value of \( c \) such that \( f'(c) = 0 \), we first need to compute the derivative of the given function \( f(x) = e^{x^3} \). Using the chain rule, the derivative is calculated as follows: \[ f'(x) = e^{x^3} \cdot (3x^2) = 3x^2 \cdot e^{x^3} \]
02

Set the Derivative Equal to Zero

Now, we need to find the value of \( c \) that makes \( f'(x) = 0 \). This results in the equation: \[ 3x^2 \cdot e^{x^3} = 0 \] Since \( e^{x^3} \) is never zero, we only need to set \( 3x^2 = 0 \).
03

Solve for x

The equation \( 3x^2 = 0 \) simplifies to \( x^2 = 0 \). Solving for \( x \) gives:\[ x = 0 \]So, \( c = 0 \).
04

Determine Local Extremum

To check if \( f(x) \) has a local extremum at \( x = 0 \), examine the second derivative test. Calculate the second derivative \( f''(x) \):\[ f''(x) = \frac{d}{dx}(3x^2 e^{x^3}) = 6x e^{x^3} + 9x^4 e^{x^3} \] Evaluating \( f''(0) \):\[ f''(0) = 6 \cdot 0 \cdot e^{0^3} + 9 \cdot 0^4 \cdot e^{0^3} = 0 \]Since both terms in \( f''(x) \) are zero at \( x = 0 \) , the second derivative test is inconclusive.
05

Conclusion on Extrema

Since the second derivative test does not help in determining the local extremum at \( x = 0 \), we will conclude based on the first derivative test.Since the function \( f(x) = e^{x^3} \) is a strictly increasing function as \( x^3 \) increases and \( e^{x^3} \) always increases over its domain, there is no local extremum at \( x = 0 \).

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Key Concepts

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

Derivative
A derivative represents the rate at which a function changes at any given point. Think of it as a tool to measure how a function is growing or shrinking at certain values of its variable. In calculus, the derivative of a function can reveal important information such as slopes of tangent lines and rates of change.
To find a derivative, we often use differentiation rules like the power rule or product rule. One such handy rule is the chain rule, essential for functions composed of other functions.
  • The process involves finding the rate of change in the outer function and multiplying it by the derivative of the inner function.
  • As an example, the derivative of the function given in our exercise, \( f(x) = e^{x^3} \), is calculated as \( f'(x) = 3x^2 \cdot e^{x^3} \).
  • This indicates that the growth of \( e^{x^3} \) is scaled by \( 3x^2 \).
Understanding derivatives is key to exploring concepts like slopes and tangent lines.
Local Extremum
A local extremum is a point where a function reaches either a minimum or maximum value within a certain neighborhood. These points are critical because they can represent peaks or valleys in the graph of a function. Finding local extrema involves locating where the derivative of the function is zero or undefined.
When we talk about finding \( c \) such that \( f'(c)=0 \), we are attempting to pinpoint these critical points.
  • In our exercise, we set \( f'(x)=0 \) which simplifies to \( 3x^2 \cdot e^{x^3} = 0 \).
  • Because \( e^{x^3} \) can never be zero, the solving simplifies further to \( x^2 = 0 \), giving us \( x = 0 \).
  • This suggests a critical point at \( x = 0 \).
However, determining whether this is a local minimum or maximum requires further tests, such as the second derivative test or observing the behavior of the function.
Chain Rule
The chain rule is a fundamental principle used to differentiate composite functions. It is crucial when dealing with functions nested within each other, allowing us to find derivatives that would otherwise seem untouchable.
Imagine a function \( g \) inside another function \( f \); the chain rule says we differentiate \( g(x) \) first, and then multiply by the derivative of \( f(g(x)) \).
  • The rule is often represented in terms of derivative notation as \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).
  • In the given exercise of \( f(x) = e^{x^3} \), \( x^3 \) is nested within the exponential function. Thus, the chain rule applies by taking \( (e^{x^3})' = e^{x^3} \cdot (3x^2) \).
  • This example clearly shows the application of the chain rule to achieve the derivative \( f'(x) = 3x^2 \cdot e^{x^3} \).
Mastering the chain rule is pivotal in advancing in calculus, especially when faced with layered functions.

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Most popular questions from this chapter

Find the equilibria of $$x_{t+1}=\sqrt{x_{t}+1}+1, \quad t=0,1,2, \ldots$$ and use the stability criterion for an equilibrium point to determine whether they are stable or unstable (you may assume that \(\left.x_{t} \geq-1\right)\)

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