Problem 21
Question
In Problems , find c such that \(f^{\prime}(c)=0\) and determine whether \(f(x)\) has a local extremum at \(x=c .\) $$ f(x)=-x^{2} $$
Step-by-Step Solution
Verified Answer
The critical point is at \( c = 0 \) and \( f(x) \) has a local maximum there.
1Step 1: Find the first derivative of f(x)
To find the critical points of the function, we need to calculate its first derivative. Given \[ f(x) = -x^2 \]The first derivative, \( f'(x) \), is calculated using the power rule:\[ f'(x) = \frac{d}{dx} (-x^2) = -2x \]
2Step 2: Find the critical point where f'(c) = 0
To find the value of \( c \) where the derivative is zero, set \[ f'(x) = 0 \]So,\[ -2x = 0 \]Solving for \( x \), we get:\[ x = 0 \]Therefore, \( c = 0 \) is the critical point where the first derivative is zero.
3Step 3: Determine if f(x) has a local extremum at x=c
To determine if there is a local extremum at \( x = c \), we investigate the second derivative, \( f''(x) \). The function \[ f''(x) = \frac{d}{dx}(-2x) = -2 \]Since \( f''(x) = -2 \) is negative, this indicates that the function is concave down at \( x = 0 \). Hence, \( f(x) \) has a local maximum at \( x = 0 \).
Key Concepts
Critical PointsDerivativeLocal Extremum
Critical Points
Critical points are essential in calculus, as they help us understand a function's behavior and identify where extreme values occur. A critical point of a function occurs where its derivative, the rate of change of the function, is equal to zero or is undefined. These points are significant because they may correspond to a local extremum or a point where the graph changes direction.
In our exercise, the given function is \[ f(x) = -x^2 \]The first step is to calculate its first derivative:\[ f'(x) = -2x \]The critical point is where this derivative equals zero: \[ -2x = 0 \]Solving for \( x \), we find \( x = 0 \). Hence, \( c = 0 \) is our critical point. This is the point where we need to analyze if there is a local extremum.
In our exercise, the given function is \[ f(x) = -x^2 \]The first step is to calculate its first derivative:\[ f'(x) = -2x \]The critical point is where this derivative equals zero: \[ -2x = 0 \]Solving for \( x \), we find \( x = 0 \). Hence, \( c = 0 \) is our critical point. This is the point where we need to analyze if there is a local extremum.
Derivative
The derivative of a function provides insight into the rate at which the function's value changes concerning its input. It is a fundamental tool in calculus used for finding slopes of tangent lines, rates of change, and solving optimization problems.
For the given function \[ f(x) = -x^2 \]its first derivative is:\[ f'(x) = -2x \]This derivative tells us how the function \( f(x) \) changes for small changes in \( x \). When the derivative equals zero, it indicates no change at that particular point, suggesting a possible extremum. Calculating derivatives is a crucial step in finding critical points and understanding the behavior of functions.
For the given function \[ f(x) = -x^2 \]its first derivative is:\[ f'(x) = -2x \]This derivative tells us how the function \( f(x) \) changes for small changes in \( x \). When the derivative equals zero, it indicates no change at that particular point, suggesting a possible extremum. Calculating derivatives is a crucial step in finding critical points and understanding the behavior of functions.
Local Extremum
A local extremum refers to points in a function where the function reaches a local maximum or minimum value. It is determined by analysis of the critical points and the concavity of the function near these points.
At the critical point \( x = 0 \) of the function\[ f(x) = -x^2 \]we calculated \[ f'(x) = -2x \]and found \( x = 0 \), let's explore the second derivative, \[ f''(x) = -2 \]which remains constant and negative. A negative second derivative indicates that the function is concave down about that point. This curvature indicates a local maximum at \( x = 0 \). Thus, at the critical point \( x = 0 \), the function achieves a local extremum, specifically a local maximum.
At the critical point \( x = 0 \) of the function\[ f(x) = -x^2 \]we calculated \[ f'(x) = -2x \]and found \( x = 0 \), let's explore the second derivative, \[ f''(x) = -2 \]which remains constant and negative. A negative second derivative indicates that the function is concave down about that point. This curvature indicates a local maximum at \( x = 0 \). Thus, at the critical point \( x = 0 \), the function achieves a local extremum, specifically a local maximum.
Other exercises in this chapter
Problem 21
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