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For each function, find all critical numbers and then use the second- derivative test to determine whether the function has a relative maximum or minimum at each critical number. $$ f(x)=x^{4}-2 x^{2}+1 $$

Short Answer

Expert verified
Critical numbers are \( x = 0, 1, -1 \). Maximum at \( x = 0 \); Minima at \( x = 1 \) and \( x = -1 \).

Step by step solution

01

Find the First Derivative

To find the critical numbers, we first need the first derivative of the function. The function is \( f(x) = x^4 - 2x^2 + 1 \). Let's find \( f'(x) \).\[ f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(2x^2) + \frac{d}{dx}(1) \]\[ f'(x) = 4x^3 - 4x \]
02

Find Critical Numbers

Critical numbers are found where \( f'(x) = 0 \) or where \( f'(x) \) is undefined. Set \( f'(x) = 0 \):\[ 4x^3 - 4x = 0 \]Factor the equation:\[ 4x(x^2 - 1) = 0 \]\[ 4x(x-1)(x+1) = 0 \]The solutions are \( x = 0, 1, -1 \). These are the critical numbers.
03

Find the Second Derivative

We use the second derivative to apply the second-derivative test. Let's find the second derivative \( f''(x) \).\[ f''(x) = \frac{d}{dx}(4x^3 - 4x) \]\[ f''(x) = 12x^2 - 4 \]
04

Apply the Second Derivative Test

Evaluate \( f''(x) \) at each critical number to determine concavity:- At \( x = 0 \): \[ f''(0) = 12(0)^2 - 4 = -4 \] (Concave Down, Maximum)- At \( x = 1 \): \[ f''(1) = 12(1)^2 - 4 = 8 \] (Concave Up, Minimum)- At \( x = -1 \): \[ f''(-1) = 12(-1)^2 - 4 = 8 \] (Concave Up, Minimum)
05

Summarize the Results

- At \( x = 0 \), there is a relative maximum.- At \( x = 1 \) and \( x = -1 \), there are relative minima.

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Key Concepts

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

Second-Derivative Test
The Second-Derivative Test is a useful tool for determining the behavior of a function at its critical numbers. Once you've found a critical number where the first derivative is zero or undefined, the second derivative can help discern whether this point is a relative maximum or minimum. Here's how it works:

  • If the second derivative at a critical number is greater than zero, the function is concave up at that point, indicating a relative minimum.
  • If the second derivative is less than zero, the function is concave down, suggesting a relative maximum.
  • If the second derivative is zero, the test is inconclusive. In such cases, other methods such as the first-derivative test might be needed to draw a conclusion.

Applying this to our exercise, we found the second derivative of the given function to be \( f''(x) = 12x^2 - 4 \). This second derivative was used to test each critical number and determine the type of extremum present at those points.
Relative Maximum
A relative maximum refers to a point on a function where its value is higher than all nearby points, effectively forming a peak.

In the example problem, we identified one such relative maximum at the critical number \( x = 0 \). The second derivative at this point, \( f''(0) = -4 \), is less than zero, indicating concavity downwards. This downward bowl shape confirms that the function peaks at this critical point, resulting in a local maximum.

Understanding where relative maxima occur provides insight into the behavioral peaks of the function, which is essential for graphing and optimization problems.
Relative Minimum
Conversely, a relative minimum is characterized by a point where the function dips lower than all nearby points, forming a valley.

In the given problem, the critical numbers \( x = 1 \) and \( x = -1 \) were identified as having relative minima. For these points, the second derivative, \( f''(1) = 8 \) and \( f''(-1) = 8 \), were both positive. This positive value tells us that the function is concave up at these points, forming a cup-like shape indicative of a local minimum.

Recognizing and understanding relative minima is crucial for tasks that involve finding the lowest possible values in a set interval, important for applications such as cost minimization or efficiency maximization.

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