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In each of Exercises \(7-22,\) use Fermat's Theorem to locate each \(c\) for which \(f(c)\) is a candidate extreme value of the given function \(f\) $$ f(x)=2 x^{2}-24 x+36 $$

Short Answer

Expert verified
The candidate extreme value is at \(c = 6\).

Step by step solution

01

Understanding Fermat's Theorem

Fermat's Theorem states that if a function has a local extreme value at a point that is within the domain and differentiable, then its derivative must be zero at that point. In other words, if there's a local maximum or minimum at point \(c\), then \(f'(c) = 0\). This means we need to differentiate the given function suiting the criteria of the theorem.
02

Differentiate the Function

Differentiate \(f(x) = 2x^2 - 24x + 36\). The derivative \(f'(x)\) is obtained by applying the power rule: \[ f'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}(24x) + \frac{d}{dx}(36) \]\[ f'(x) = 4x - 24 \] Constant terms differentiate to zero.
03

Set the Derivative Equal to Zero

To find the critical points, set the derivative equal to zero: \[ 4x - 24 = 0 \]Solve for \(x\).
04

Solve for Critical Point \(c\)

Add 24 to both sides of the equation to solve for \(x\):\[ 4x = 24 \]Divide both sides by 4:\[ x = 6 \] Thus, the critical point is \(c = 6\).

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Key Concepts

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

Critical Points
In calculus, a critical point of a function is where the derivative of the function is either zero or undefined. This concept is fundamental when analyzing the behavior of functions, primarily when you're interested in identifying locations where a function might have extreme values, such as its maxima or minima.

Let's unpack this in the context of the function given, which is \( f(x) = 2x^2 - 24x + 36 \). The derivative, \( f'(x) = 4x - 24 \), gives us an equation that helps us pinpoint those critical points. To locate them, you typically set \( f'(x) = 0 \) and solve for \( x \). In our case, setting \( 4x - 24 = 0 \) and solving gives \( x = 6 \). This means that \( x = 6 \) is a critical point.

The critical points are crucial because they are where potential extreme values of the function can occur. Without identifying these, we'd miss out on important aspects of the function's behavior.
Differentiation
Differentiation is a process in calculus that involves finding the derivative of a function. The derivative represents the rate at which a function's value changes. Understanding this concept is central to identifying critical points and, by extension, extreme values of a function.

For our given function \( f(x) = 2x^2 - 24x + 36 \), finding the derivative involves applying the power rule. The power rule is one of the basic techniques in differentiation, where the derivative of \( ax^n \) is \( nax^{n-1} \). Applying this to each part of the function:
  • The derivative of \( 2x^2 \) is \( 4x \).
  • The derivative of \(-24x \) is \(-24 \).
  • The derivative of the constant \(36\) is \(0\), since constants have no rate of change.

Thus, we find that \( f'(x) = 4x - 24 \). This simple linear equation leads us directly to the critical point, where the derivative equates to zero. Differentiation allows us to better understand the shape and behavior of functions, allowing for a mathematical interpretation of real-world changes.
Extreme Values
Extreme values of a function are its local maxima and minima, critical for understanding the function's overall behavior. They are vital in various applications, from optimizing resources to calculating maximum potential outcomes in both theoretical and practical scenarios.

Applying Fermat's Theorem gives us a straightforward path to identifying candidate points for these extremes. Fermat's Theorem assures us that if the function has a local extremum and is differentiable at a point, the derivative is zero. Once we determine the critical points, such as \( x = 6 \) for \( f(x) = 2x^2 - 24x + 36 \), we can further analyze if these points are indeed locations of extreme values.

Finding extreme values isn't just about finding where the derivative is zero, but understanding the overall function. The context of the problem could require checking the behavior of the function around these points with a second derivative test or an analysis of the first derivative's sign changes. This confirms whether a critical point is a minimum, maximum, or a saddle point. Recognizing these extremes allows us to grasp the full operational capacity of a given function in diverse scenarios.

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Most popular questions from this chapter

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