Problem 14 - Calculus Single Variable

Question

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 \(f\) $$ f(x)=x+1 / x $$

Step-by-Step Solution

Answer

Critical points where \( f(c) \) could be an extreme value are \( c = 1 \) and \( c = -1 \).

Step 1: Understand the function

Our task is to find the critical points of the function \( f(x) = x + \frac{1}{x} \) using Fermat's Theorem. Fermat's Theorem suggests that if \( f \) has a local maximum or minimum at \( c \), and if the derivative \( f'(c) \) exists, then \( f'(c) = 0 \).

Step 2: Differentiate the function

Differentiate the function \( f(x) = x + \frac{1}{x} \) to find \( f'(x) \). The derivative is:\[f'(x) = 1 - \frac{1}{x^2}\]This is obtained by differentiating \( x \) to get 1, and \( \frac{1}{x} \) to get \(-\frac{1}{x^2}\).

Step 3: Find critical points

Set the derivative \( f'(x) = 1 - \frac{1}{x^2} \) equal to zero to solve for \( x \):\[1 - \frac{1}{x^2} = 0\]Rearrange to get:\[\frac{1}{x^2} = 1\]Taking the reciprocal gives:\[x^2 = 1\]This results in two solutions: \( x = 1 \) and \( x = -1 \).

Step 4: Verify critical points

Check that \( x = 1 \) and \( x = -1 \) are within the domain of \( f \). Since \( f(x) = x + \frac{1}{x} \) is defined for all \( x eq 0 \), both \( x = 1 \) and \( x = -1 \) are valid critical points.

Key Concepts

Critical Points

In calculus, critical points of a function are significant because they help us locate potential places where the function reaches a local maximum or minimum. A critical point occurs where the derivative of the function is zero or undefined.

• To find these points, we first differentiate the function.
• Then, we set the derivative equal to zero.

Using Fermat's Theorem, we recognize that critical points are candidates for extreme values. For the function \(f(x) = x + \frac{1}{x}\), the derivative is \(f'(x) = 1 - \frac{1}{x^2}\). Setting this derivative to zero gives us:
\[1 - \frac{1}{x^2} = 0\]This simplifies to \[x^2 = 1\], resulting in critical points at \(x = 1\) and \(x = -1\). These are the potential points where the function can have extreme values.

Differentiation

Differentiation is a fundamental concept in calculus. It refers to the process of calculating the derivative of a function, which represents the rate at which the function's value changes as its input changes.When we differentiate \(f(x) = x + \frac{1}{x}\), we handle each term separately.

• The differentiation of \(x\) with respect to \(x\) is simply 1.
• The term \(\frac{1}{x}\) is a bit trickier, but it differentiates to \(-\frac{1}{x^2}\).

Combining these results, the derivative \(f'(x)\) becomes \(1 - \frac{1}{x^2}\). This derivative is instrumental in finding critical points and analyzing the behavior of \(f(x)\). It tells us when the function is increasing, decreasing, or possibly achieving extreme values.

Extreme Values

Extreme values of a function are the largest or smallest values in its range within a given interval. These can be local maxima or minima, which occur at the critical points, and they offer insights into the function’s overall behavior.To identify whether the critical points correspond to extreme values, additional analysis, such as the second derivative test or the first derivative test, can be applied. For \(f(x) = x + \frac{1}{x}\), we've located critical points at \(x = 1\) and \(x = -1\).

• At these points, observe the function's behavior to determine if they are indeed maximum or minimum points.
• If the function changes from increasing to decreasing at a critical point, you've found a local maximum.
• Similarly, if it changes from decreasing to increasing, it signifies a local minimum.

In conjunction, understanding extreme values and their conditions is essential for getting a complete picture of a function's profile.