Problem 109
Question
Show that the quadratic function $$ f(x)=a x^{2}+b x+c \quad(a \neq 0) $$ has a relative extremum when \(x=-b / 2 a\). Also, show that the relative extremum is a relative maximum if \(a<0\) and a relative minimum if \(a>0\).
Step-by-Step Solution
Verified Answer
The quadratic function \(f(x) = ax^2 + bx + c\) has a relative extremum at \(x = -\frac{b}{2a}\). The extremum is a relative maximum if \(a < 0\) and a relative minimum if \(a > 0\).
1Step 1: Compute the derivative
First, we'll compute the derivative of the function \(f(x)\).
Given:
\[f(x) = ax^2 + bx + c\]
The derivative of the function is obtained as follows:
\[f'(x) = \frac{d}{dx}(ax^2 + bx + c)\]
Now, we can apply the power rule to find the derivative:
\[f'(x) = 2ax + b\]
2Step 2: Set the derivative equal to 0
We need to find the critical points by setting the derivative equal to 0.
\[2ax + b = 0\]
Now, we need to solve for x:
\[x = -\frac{b}{2a}\]
3Step 3: Find the concavity using the second derivative
To find the concavity of the critical point, we need to compute the second derivative of the function. Since \(f'(x) = 2ax + b\), let's find the second derivative \(f''(x)\):
\[f''(x) = \frac{d}{dx}(2ax + b)\]
Applying the derivative rules, we find:
\[f''(x) = 2a\]
The concavity of the function at the point x depends on the sign of the second derivative.
4Step 4: Determine if the extremum is a minimum or maximum
Now, we can determine if the extremum is a relative minimum or a relative maximum.
• If \(a > 0\), then the second derivative \(f''(x) > 0\), which indicates that the function is concave up at the critical point. Therefore, the extremum is a relative minimum.
• If \(a < 0\), then the second derivative \(f''(x) < 0\), which indicates that the function is concave down at the critical point. Therefore, the extremum is a relative maximum.
In conclusion, the quadratic function \(f(x) = ax^2 + bx + c\) has a relative extremum at \(x = -\frac{b}{2a}\). The extremum is a relative maximum if \(a < 0\) and a relative minimum if \(a > 0\).
Key Concepts
DerivativesCritical PointsConcavityRelative Extrema
Derivatives
Derivatives are fundamental to understanding how functions behave. In the context of a quadratic function, the derivative helps us find the slope of the tangent line at any given point on the graph. For the quadratic function \[f(x) = ax^2 + bx + c\],we calculated the derivative as \[f'(x) = 2ax + b\].This formula tells us the rate of change of the function with respect to \(x\). By setting this derivative equal to zero, we can identify critical points where the slope of the tangent is horizontal. Such points are important because they often correspond to maximum or minimum values, which are the relative extrema of the function.
Critical Points
Critical points are values of \(x\) where the derivative of a function is zero or undefined, indicating potential points of interest such as maxima or minima. For the quadratic function, we found the critical point by solving \[2ax + b = 0\], which results in \[ x = -\frac{b}{2a} \]. This critical point is where the slope of the tangent line is zero, implying that the function changes direction. It is at these critical points that a function could have a peak or a trough, which are key to understanding the function's behavior on a graph.
Concavity
Concavity describes the direction the parabola opens and helps determine the nature of the critical points. The second derivative of a function gives us insight into its concavity. For the quadratic function, the second derivative is \[f''(x) = 2a\]. Depending on the sign of \(a\):
- If \(a > 0\), then \(f''(x) > 0\), indicating the function is concave up, like a cup facing upwards.
- If \(a < 0\), then \(f''(x) < 0\), indicating the function is concave down, like a cap.
Relative Extrema
The terms relative minimum and relative maximum refer to points where a function reaches its lowest or highest local values, respectively. In a quadratic function:
- When \(a > 0\) (the parabola opens upward), the critical point \(x = -\frac{b}{2a}\) is a relative minimum because the function's graph has a trough.
- When \(a < 0\) (the parabola opens downward), the critical point is a relative maximum because the graph peaks at this point.
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