Finding Local Maxima And Minima Of A Function

by Alex Johnson 46 views

Let's dive into the fascinating world of calculus to uncover the local extrema of the function $f(x) = 2 + 2x + 50x^{-1}$. Understanding where a function reaches its peaks (local maxima) and valleys (local minima) is a fundamental concept in calculus, providing crucial insights into the behavior of functions. This function, though seemingly simple, presents a great opportunity to practice these essential techniques. We're going to embark on a journey to find these critical points, determine whether they represent a maximum or minimum, and calculate the corresponding values. So, grab your calculus toolkit, and let's get started!

Understanding Local Extrema

Before we start crunching numbers, let's clarify what we mean by local minimum and local maximum. Imagine a mountain range; a local maximum is like the peak of a smaller hill, not necessarily the highest mountain in the entire range, but certainly higher than all the points immediately surrounding it. Similarly, a local minimum is like the bottom of a small valley, lower than all the points nearby. Mathematically, a function $f(x)$ has a local maximum at a point $c$ if $f(c) gtr f(x)$ for all $x$ in some open interval containing $c$. Conversely, $f(x)$ has a local minimum at $c$ if $f(c) < f(x)$ for all $x$ in some open interval containing $c$. These points are crucial for understanding the shape and behavior of a function's graph, helping us identify regions of increase and decrease, and pinpointing specific turning points. The first derivative of a function plays a starring role in this quest. Where the derivative is zero or undefined, we find critical points. These critical points are our candidates for local maxima and minima. The second derivative test offers a powerful way to classify these critical points without needing to analyze the function's behavior in intervals.

Finding Critical Points

Our first major step in finding the local extrema of $f(x) = 2 + 2x + 50x^{-1}$ is to locate its critical points. Critical points are the locations where the derivative of the function, $f'(x)$, is either equal to zero or undefined. Let's begin by finding the derivative of our function. Using the power rule for differentiation, which states that the derivative of $ax^n$ is $anx^{n-1}$, we can differentiate each term of $f(x)$:

  • The derivative of the constant term '2' is 0.
  • The derivative of $2x$ (where $n=1$) is $2 imes 1 imes x^{1-1} = 2x^0 = 2$.
  • The derivative of $50x^{-1}$ (where $n=-1$) is $50 imes (-1) imes x^{-1-1} = -50x^{-2}$.

So, the derivative of our function is $f'(x) = 0 + 2 - 50x^{-2}$, which simplifies to $f'(x) = 2 - rac{50}{x^2}$.

Now, we need to find the values of $x$ for which $f'(x) = 0$ or $f'(x)$ is undefined.

Setting $f'(x) = 0$:

$2 - rac{50}{x^2} = 0$

To solve for $x$, we can add $ rac{50}{x^2}$ to both sides:

$2 = rac{50}{x^2}$

Multiply both sides by $x^2$ (assuming $x eq 0$):

$2x^2 = 50$

Divide by 2:

$x^2 = 25$

Taking the square root of both sides gives us two possible values for $x$:

$x = pm{5} ext{ and } x = pm{-5}$

Where $f'(x)$ is undefined:

The derivative $f'(x) = 2 - rac{50}{x^2}$ is undefined when the denominator, $x^2$, is equal to zero. This occurs when $x = 0$. However, we must also consider the domain of the original function $f(x) = 2 + 2x + 50x^{-1}$. The term $50x^{-1}$ means $ rac{50}{x}$, which is undefined at $x=0$. Therefore, $x=0$ is not in the domain of $f(x)$ and cannot be a critical point for local extrema.

Thus, our critical points are $x = 5$ and $x = -5$. These are the potential locations for our local maximum and local minimum.

Using the Second Derivative Test

To determine whether our critical points $x = 5$ and $x = -5$ correspond to a local maximum or a local minimum, we can employ the Second Derivative Test. This test involves calculating the second derivative of our function, $f''(x)$, and then evaluating it at each critical point. The second derivative tells us about the concavity of the function: if $f''(c) > 0$, the function is concave up at $c$, indicating a local minimum; if $f''(c) < 0$, the function is concave down at $c$, indicating a local maximum; and if $f''(c) = 0$, the test is inconclusive, and we might need to use the first derivative test.

Let's find the second derivative, $f''(x)$. We start with our first derivative: $f'(x) = 2 - 50x^{-2}$. Now, we differentiate this again using the power rule:

  • The derivative of the constant term '2' is 0.
  • The derivative of $-50x^{-2}$ (where $n=-2$) is $-50 imes (-2) imes x^{-2-1} = 100x^{-3}$.

So, the second derivative is $f''(x) = 0 + 100x^{-3}$, which simplifies to $f''(x) = rac{100}{x^3}$.

Now, we evaluate $f''(x)$ at our critical points:

At $x = 5$:

$f''(5) = rac{100}{5^3} = rac{100}{125}$

Since $ rac{100}{125} > 0$, the function is concave up at $x = 5$. According to the Second Derivative Test, this means there is a local minimum at $x = 5$.

At $x = -5$:

$f''(-5) = rac{100}{(-5)^3} = rac{100}{-125}$

Since $ rac{100}{-125} < 0$, the function is concave down at $x = -5$. This indicates that there is a local maximum at $x = -5$.

Calculating the Values of the Extrema

We've successfully identified the locations of our local maximum and minimum. Now, the final step is to calculate the actual values of the function at these points. This is done by substituting the x-values of our critical points back into the original function, $f(x) = 2 + 2x + 50x^{-1}$.

Value at the local minimum ($x = 5$):

$f(5) = 2 + 2(5) + rac{50}{5}$

$f(5) = 2 + 10 + 10$

$f(5) = 22$

So, the local minimum occurs at $x = 5$ with a value of 22. This means the lowest point in the vicinity of $x=5$ on the graph of $f(x)$ is at the coordinates $(5, 22)$.

Value at the local maximum ($x = -5$):

$f(-5) = 2 + 2(-5) + rac{50}{-5}$

$f(-5) = 2 - 10 - 10$

$f(-5) = -18$

Therefore, the local maximum occurs at $x = -5$ with a value of -18. This means the highest point in the vicinity of $x=-5$ on the graph of $f(x)$ is at the coordinates $(-5, -18)$.

Conclusion

In conclusion, by applying the principles of differential calculus, we have successfully analyzed the function $f(x) = 2 + 2x + 50x^{-1}$. We first found the critical points by setting the first derivative, $f'(x) = 2 - rac{50}{x^2}$, to zero, which yielded $x = 5$ and $x = -5$. Subsequently, we utilized the Second Derivative Test, calculating $f''(x) = rac{100}{x^3}$. Evaluating the second derivative at our critical points revealed that $f''(5) > 0$, indicating a local minimum at $x = 5$, and $f''(-5) < 0$, indicating a local maximum at $x = -5$. Finally, substituting these x-values back into the original function, we found the value of the local minimum to be $f(5) = 22$ and the value of the local maximum to be $f(-5) = -18$. This thorough examination provides a complete picture of the turning points of this function.

Understanding these concepts is vital for various applications in mathematics, science, and engineering, from optimization problems to curve sketching. For further exploration into calculus and function analysis, you might find the resources at **Khan Academy** helpful. They offer a wealth of free lessons and practice problems covering these topics and much more. Another excellent resource for deeper mathematical understanding is **Brilliant.org**, which provides interactive courses and challenges.