Chapter 3: Problem 1
Using elementary calculus, show that the minimum and maximum points for \(y=f(x)\) occur among the minimum and maximum points for \(y=f^{2}(x)\). Assuming \(f(x) \geq 0\), why can we minimize \(f(x)\) by minimizing \(f^{2}(x)\) ?
Short Answer
Expert verified
Minimizing \( f^2(x) \) corresponds to minimizing \( f(x) \) for \( f(x) \geq 0 \), as their extremal points coincide.
Step by step solution
01
Define functions and constraints
We have two functions: \( y = f(x) \) and \( y = f^2(x) \). Given the condition \( f(x) \geq 0 \), when \( f(x) = 0 \), \( f^2(x) \) also equals 0. Our goal is to show that the extremal points of \( f(x) \) correspond to the extremal points of \( f^2(x) \).
02
Differentiate both functions
To find the critical points, differentiate both functions with respect to \( x \). For the first function, the derivative is \( f'(x) \). For the second function, use the chain rule to differentiate: \[ \frac{d}{dx}[f^2(x)] = 2f(x)f'(x). \]
03
Find critical points of \( y = f(x) \)
Set \( f'(x) = 0 \). The points where \( f'(x) = 0 \) are the critical points for \( y = f(x) \). These are potential minima, maxima, or saddle points.
04
Find critical points of \( y = f^2(x) \)
Set \( 2f(x)f'(x) = 0 \). This equation is satisfied when \( f(x) = 0 \) or \( f'(x) = 0 \). Since the condition \( f(x) \geq 0 \) is given, it implies that the critical points from this condition are also the critical points for \( y = f(x) \).
05
Analyze the common critical points
The critical points of \( y = f(x) \) at \( f'(x) = 0 \) overlap with critical points of \( y = f^2(x) \) because both conditions include \( f'(x) = 0 \). Thus, the extremum points of \( y = f(x) \) occur among those of \( y = f^2(x) \).
06
Conclusion on minimization of \( f(x) \) via \( f^2(x) \)
Since the minimum point of \( f^2(x) \) corresponds to the points where \( f(x) \) is minimized (for \( f(x) \geq 0 \)), minimizing \( f^2(x) \) will also minimize \( f(x) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
In calculus, critical points are essential in understanding the behavior of a function. A critical point of a function, say \(y = f(x)\), is where the function’s slope or derivative is zero \((f'(x) = 0)\) or undefined. These points are important because they are where the function could potentially have a local maximum or minimum, or a point of inflection.
- Local maximum: The point where a function reaches its highest value in a given interval.
- Local minimum: The point where a function attains its lowest value in a given interval.
- Point of inflection: A point on the curve where the curvature changes sign.
Differentiation
Differentiation is a fundamental tool in calculus used to determine the rate at which a quantity changes. When we differentiate a function, we're essentially discovering how the function's output changes with small changes in its input. This can reveal a lot about its behavior, such as finding critical points.
- The derivative of a function \(f(x)\) is denoted by \(f'(x)\) or \(\frac{df}{dx}\).
- The process involves rules such as the power rule, product rule, quotient rule, and chain rule.
Function Minimization
Function minimization is the process of finding the smallest value that a function can take. For the function \(y = f(x)\), minimization involves finding the value of \(x\) at which \(f(x)\) reaches its minimum. This is especially useful in various applications, from physical systems to economics.
- The process often involves locating critical points and then using tests (such as the second derivative test) to classify these points as minima or maxima.
- For \(f(x) \geq 0\), minimizing \(f(x)\) is comparable to minimizing \(f^2(x)\) since both share the same critical points when \(f(x) = 0\) or \(f'(x) = 0\).
- In particular, since the square function \(f^2(x)\) is always non-negative, it ensures that finding the minima of \(f^2(x)\) guarantees reducing \(f(x)\) to the lowest possible non-negative value.