Calculus Challenge: Differentiating And Solving For Zero
Hey everyone! Today, we're diving into a cool calculus problem. We've got a function, and we're going to find its derivative and then figure out where that derivative equals zero. This is a classic exercise that helps us understand how functions behave. Let's get started, shall we?
(a) Finding the Derivative:
Alright, first things first, we need to find the derivative of the function. Remember, the derivative tells us the instantaneous rate of change of the function at any given point. Our function is $y = 4x^2 - \frac{1}{x}$. To make things a bit easier, let's rewrite the function. Instead of $\frac{1}{x}$, we can write it as $x^{-1}$. This way, we can apply the power rule more directly. So now we have $y = 4x^2 - x^{-1}$. The power rule states that if we have a term like $ax^n$, its derivative is $nax^{n-1}$. Let's apply this to each term in our function.
For the first term, $4x^2$, the derivative is $2 * 4x^{2-1}$, which simplifies to $8x$. For the second term, $-x^{-1}$, the derivative is $-1 * -1x^{-1-1}$, which simplifies to $x^{-2}$. Remember that the derivative of a sum or difference is just the sum or difference of the derivatives. Therefore, by combining these two, we get the derivative of the original function. So, applying the power rule, the derivative, $\frac{dy}{dx}$, is $8x + x^{-2}$. We can rewrite $x^{-2}$ as $\frac{1}{x^2}$.
So, our final answer for part (a) is: $\frac{dy}{dx} = 8x + \frac{1}{x^2}$. Easy peasy, right? We've successfully found the derivative of our function. Now that we've found the derivative, we know how the function is changing at any point. We can use this to find the critical points, where the function might have a maximum or minimum value. This is super useful in optimization problems, where we are trying to find the best possible outcome.
In essence, finding the derivative is like zooming in on the function at any point and seeing its slope. A positive slope means the function is increasing, a negative slope means it's decreasing, and a slope of zero means it's momentarily flat β a potential turning point. By finding where the derivative is zero, we pinpoint these turning points, which are crucial in understanding the overall shape and behavior of the function. Think of it as finding the function's sweet spots, where it might reach its highest or lowest value. Understanding derivatives unlocks a deeper understanding of how things change and allows us to make predictions and solve problems in various fields, from physics and engineering to economics and beyond.
Now, let's move on to the next part of the problem. We're going to use the derivative we just found to solve something even more interesting. Are you ready?
(b) Solving
Okay, time for the fun part! Now that we have the derivative, $\fracdy}{dx} = 8x + \frac{1}{x^2}$, we need to find the values of x for which the derivative equals zero. This is where things get interesting. Setting the derivative to zero allows us to find the critical points of the original function. These are the points where the function might have a maximum, a minimum, or a saddle point. Let's set up the equation{x^2} = 0$. Our goal is to isolate x and solve for its value. First, let's get rid of that fraction. We can do this by multiplying both sides of the equation by $x^2$. This gives us: $8x^3 + 1 = 0$.
Now, let's rearrange the equation to isolate the term with x. Subtract 1 from both sides: $8x^3 = -1$. Next, we'll divide both sides by 8: $x^3 = -\frac1}{8}$. To solve for x, we need to take the cube root of both sides{8}}$. The cube root of -1/8 is -1/2. So, we get $x = -\frac{1}{2}$. This value of x is a critical point of our original function. It means that at $x = -\frac{1}{2}$, the slope of the original function is zero. This point could potentially be a local maximum, a local minimum, or a point of inflection.
To determine the nature of this critical point (maximum, minimum, or neither), we could use the second derivative test. The second derivative tells us about the concavity of the function. If the second derivative is positive at the critical point, it's a local minimum; if it's negative, it's a local maximum; and if it's zero, the test is inconclusive. While the question doesn't ask us to find the nature of the critical point, understanding this concept is crucial in calculus. Let's quickly find the second derivative of the original function and find its concavity. We have to differentiate the derivative, $\frac{dy}{dx} = 8x + \frac{1}{x^2} = 8x + x^{-2}$. Then, the second derivative, $\frac{d2y}{dx2} = 8 - 2x^{-3} = 8 - \frac{2}{x^3}$. At $x = -\frac{1}{2}$, the second derivative equals $8 - \frac{2}{(-\frac{1}{2})^3} = 8 - \frac{2}{-\frac{1}{8}} = 8 - (-16) = 24$. Since the second derivative is positive at $x = -\frac{1}{2}$, then the critical point is a local minimum. So, our function has a local minimum at $x = -\frac{1}{2}$.
Solving for when the derivative equals zero helps us to pinpoint important features of the function's graph. These are the points where the function changes direction. This is a fundamental concept in calculus, and by mastering it, you're well on your way to understanding more advanced concepts and problem-solving techniques.
Conclusion
So there you have it, folks! We've successfully found the derivative of the function, and then solved for when that derivative equals zero. This has led us to identify a critical point, which we then determined to be a local minimum. Remember, understanding derivatives and how to solve for their zeros is a powerful tool in calculus. Keep practicing, and you'll become a calculus whiz in no time!
This entire process is a cornerstone of calculus. It helps us understand the behavior of functions and solve a wide range of real-world problems. Whether you're a student learning calculus for the first time or brushing up on your skills, this problem provides a solid foundation for more complex concepts.
Feel free to experiment with different functions and try finding their derivatives and critical points. The more you practice, the better you'll become at mastering these essential calculus concepts. Keep up the great work and happy calculating!