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Find all critical values, the largest open intervals on which \(f\) is increasing, the largest open intervals on which \(f\) is decreasing, and all relative maxima and minima. Sketch a rough graph of \(f\). In Exercises 37 through 42, assume that the constants \(a\) and \(b\) are positive. \(f(x)=x^{4}+2 x^{2}+1\)

Short Answer

Expert verified
Critical value at \( x = 0 \). \( f \) is decreasing on \( (-\infty, 0) \) and increasing on \( (0, \infty) \). Relative minimum at \( (0, 1) \).

Step by step solution

01

Find the Derivative of \( f(x) \)

To find the critical values, first differentiate the function: \( f(x) = x^4 + 2x^2 + 1 \). The derivative is found using the power rule:\[ f'(x) = 4x^3 + 4x. \]
02

Set the Derivative Equal to Zero to Find Critical Points

Set the derivative equal to zero to find the critical points:\[ 4x^3 + 4x = 0. \]Factor out the common term:\[ 4x(x^2 + 1) = 0. \]This gives us potential critical points at \( x = 0 \). The term \( x^2 + 1 \) never equals zero since \( x^2 + 1 \geq 0 \) for all real \( x \). Thus, the only critical point is at \( x = 0 \).
03

Determine Intervals of Increase and Decrease

To determine where the function is increasing or decreasing, test the sign of \( f'(x) = 4x(x^2 + 1) \) around the critical point. Choose test points such as \( x = -1 \) and \( x = 1 \).- For \( x < 0 \) (e.g., \( x = -1 \)), \( f'(-1) = 4(-1)((-1)^2 + 1) = -8 \) (negative), so \( f \) is decreasing.- For \( x > 0 \) (e.g., \( x = 1 \)), \( f'(1) = 4(1)(1^2 + 1) = 8 \) (positive), so \( f \) is increasing.
04

Identify Relative Maxima and Minima

At the critical point \( x = 0 \), the derivative changes from negative to positive, indicating a relative minimum. Therefore, \( f \) has a relative minimum at \( x = 0 \).Evaluate \( f(0) \):\[ f(0) = 0^4 + 2(0^2) + 1 = 1. \]Thus, there is a relative minimum at \( (0, 1) \).
05

Sketch a Rough Graph of \( f(x) \)

Based on the derivative test:- The function decreases on the interval \( (-\infty, 0) \).- The function increases on the interval \( (0, \infty) \).The graph is a smooth, continuous curve starting from positive infinity on the left end, decreasing toward the relative minimum at \( (0, 1) \), then increasing to positive infinity on the right end.

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Key Concepts

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

Derivative Test
The derivative test is an essential tool in calculus to find critical points, which are potential locations of relative extrema (maximum or minimum values) on a function. To perform this test, we start by finding the derivative of the function, which represents the rate of change of the function with respect to its variable. In our example, the function is given by:
\[ f(x) = x^4 + 2x^2 + 1. \]
Applying the power rule for differentiation, we compute the derivative as:
\[ f'(x) = 4x^3 + 4x. \]
The next step is to set this derivative equal to zero in order to find the critical points, as these are where the slope of the tangent to the curve is horizontal, potentially indicating peaks or troughs in the graph. Solving:
\[ 4x( x^2 + 1) = 0, \]reveals that the critical point is at \( x = 0 \), because \( x^2 + 1 \geq 0 \) does not produce additional solutions for real \( x \). This process of using the derivative to identify points where the function's behavior changes is known as the derivative test.
Relative Extrema
Relative extrema refer to points in the domain of a function where it reaches local maximum or minimum values. Once the critical points are found using the derivative test, we determine the behavior of the function around these points. This helps in identifying whether each critical point is a location of a relative maximum or minimum.
For our function \( f(x) = x^4 + 2x^2 + 1 \), we identified the critical point at \( x = 0 \). To ascertain whether this is a maximum or minimum, we need to analyze the sign change of the derivative around this point.
  • Before \( x = 0 \): Choosing a point like \( x = -1 \), we find \( f'(-1) = -8 \), indicating the function is decreasing.
  • After \( x = 0 \): Choosing a point like \( x = 1 \), we find \( f'(1) = 8 \), showing the function is increasing.
The change from decreasing to increasing at \( x = 0 \) indicates a relative minimum. Evaluating the function at this point gives us \( f(0) = 1 \), confirming a relative minimum at the point \( (0, 1) \). This analysis helps us understand critical behaviors of the function's graph, particularly around the extrema.
Interval of Increase and Decrease
Understanding where a function is increasing or decreasing is crucial for graphing and applications of calculus. These intervals can be determined using the first derivative test, which tells us the sign of the derivative across intervals divided by critical points.
For the function \( f(x) = x^4 + 2x^2 + 1 \), we calculated its derivative as \( f'(x) = 4x(x^2 + 1) \). Testing this derivative around the critical point \( x=0 \) gives us the intervals of increase and decrease. Here's the strategy:
  • If \( f'(x) > 0 \), the function is increasing.
  • If \( f'(x) < 0 \), the function is decreasing.
From our earlier analysis:
  • For \( x < 0 \), such as at \( x = -1 \), \( f'(-1) = -8 \). Hence, \( f(x) \) is decreasing on \((- \infty, 0)\).
  • For \( x > 0 \), such as at \( x = 1 \), \( f'(1) = 8 \). Hence, \( f(x) \) is increasing on \((0, \infty)\).
These intervals reveal how the function behaves across its domain: decreasing towards the relative minimum at \( x = 0 \) and then increasing onward, giving us a clear view of the function's graph.

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