Find F'(4) For F(x) = √(5x + 5): Step-by-Step Solution
Hey math enthusiasts! Today, we're diving into a calculus problem that might seem a bit daunting at first, but trust me, we'll break it down together. We're going to figure out how to find the derivative, f'(a), of the function f(x) = √(5x + 5) at the specific point a = 4. So, grab your calculators, and let's get started!
Understanding the Problem: Derivatives and All That Jazz
Before we jump into the nitty-gritty, let's make sure we're all on the same page. What exactly are we trying to find? Well, f'(a) represents the derivative of the function f(x) evaluated at the point x = a. In simpler terms, it tells us the instantaneous rate of change of the function at that particular point. Think of it as the slope of the tangent line to the curve of the function at x = a. Why is this important? Derivatives are fundamental to calculus and have tons of applications in physics, engineering, economics, and pretty much any field that deals with rates of change. Whether it's optimizing designs, understanding motion, or predicting market trends, derivatives are our go-to tool.
Now, for our specific problem, we have the function f(x) = √(5x + 5), which is a square root function. We want to find its derivative at a = 4. This means we need to find how the function is changing at the point where x is equal to 4. We'll use the definition of the derivative to solve this, and I'll walk you through each step. So, don’t worry if it seems complicated now; by the end of this, you'll be a pro at finding derivatives of square root functions!
Step 1: The Definition of the Derivative
Okay, first things first, we need to remember the definition of the derivative. This is the foundation of everything we're doing, so let's get it crystal clear. The derivative of a function f(x) at a point x = a, denoted as f'(a), is defined as the limit:
f'(a) = lim (h→0) [f(a + h) - f(a)] / h
This might look a bit scary, but don't fret! Let’s break it down. The lim (h→0) part means we're looking at what happens as h gets incredibly close to zero. The h here is a tiny change in x. We're essentially looking at what happens to the function as we zoom in closer and closer to the point x = a. The expression [f(a + h) - f(a)] / h is the difference quotient, which represents the average rate of change of the function over a small interval. As h approaches zero, this average rate of change becomes the instantaneous rate of change, which is exactly what the derivative is.
So, in plain English, the derivative is the limit of the difference quotient as the change in x approaches zero. This definition is super important because it gives us a way to calculate the derivative from first principles. We're not just using some magic formula; we're understanding the fundamental concept of how a function changes at a specific point. Now that we have this definition in our toolkit, we're ready to tackle our specific problem. Let's plug in our function and value of a and see what happens!
Step 2: Plugging in the Values
Alright, let's get our hands dirty and plug in the values into the definition of the derivative. We know that f(x) = √(5x + 5) and a = 4. So, we need to find f(a + h) and f(a). First, let's find f(a). This is the easiest part. We just substitute a = 4 into our function:
f(4) = √(5(4) + 5) = √(20 + 5) = √25 = 5
So, f(4) = 5. Easy peasy! Now, let's tackle f(a + h). This is where we substitute (a + h) into our function. Since a = 4, we have:
f(4 + h) = √(5(4 + h) + 5)
Let’s simplify this a bit:
f(4 + h) = √(20 + 5h + 5) = √(25 + 5h)
Okay, we've got f(4 + h) = √(25 + 5h). Now we have all the pieces we need to plug into the definition of the derivative. Remember, the definition is:
f'(a) = lim (h→0) [f(a + h) - f(a)] / h
Substituting our values, we get:
f'(4) = lim (h→0) [√(25 + 5h) - 5] / h
This looks a bit more complicated, but we're making progress! We've successfully plugged in our function and the value of a into the definition of the derivative. Now, we need to figure out how to evaluate this limit. If we try to directly substitute h = 0, we get (√25 - 5) / 0 = 0 / 0, which is an indeterminate form. This means we can't just plug in h = 0; we need to do some algebraic magic to simplify the expression. That's where our next step comes in: rationalizing the numerator. Stay tuned!
Step 3: Rationalizing the Numerator
Alright, we've hit a bit of a snag. We can't just plug in h = 0 into our limit expression because we get an indeterminate form. But don't worry, this is a common situation in calculus, and we have a trick up our sleeve: rationalizing the numerator. What does that mean? Well, it means we're going to multiply the numerator and the denominator of our fraction by the conjugate of the numerator. The conjugate of √(25 + 5h) - 5 is √(25 + 5h) + 5. We change the sign in the middle, and that's it!
So, we're going to multiply both the numerator and the denominator by √(25 + 5h) + 5. This might seem a bit random, but trust me, it's going to help us simplify the expression. Here's what it looks like:
f'(4) = lim (h→0) [√(25 + 5h) - 5] / h * [√(25 + 5h) + 5] / [√(25 + 5h) + 5]
Now, let's multiply the numerators. Remember the difference of squares formula: (a - b)(a + b) = a² - b². This is exactly what we have in our numerator. So, we get:
[√(25 + 5h) - 5][√(25 + 5h) + 5] = (√(25 + 5h))² - 5² = (25 + 5h) - 25 = 5h
Cool, the numerator simplifies to 5h! Now let’s rewrite our limit expression:
f'(4) = lim (h→0) 5h / [h(√(25 + 5h) + 5)]
Notice anything? We have an h in both the numerator and the denominator! This means we can cancel them out:
f'(4) = lim (h→0) 5 / [√(25 + 5h) + 5]
Awesome! We've simplified the expression significantly. Now, we can try plugging in h = 0 again and see what happens.
Step 4: Evaluating the Limit
We've done some algebraic wizardry and simplified our expression quite a bit. We're now at:
f'(4) = lim (h→0) 5 / [√(25 + 5h) + 5]
Remember, the whole reason we rationalized the numerator was to get rid of that pesky indeterminate form. Let's see if it worked! Now we can try substituting h = 0 directly into the expression:
f'(4) = 5 / [√(25 + 5(0)) + 5] = 5 / [√25 + 5] = 5 / [5 + 5] = 5 / 10 = 1/2
Boom! We got a value! So, the limit exists, and it's equal to 1/2. This means:
f'(4) = 1/2
We've found the derivative of our function f(x) = √(5x + 5) at the point a = 4. It's 1/2. This tells us that at x = 4, the function is changing at a rate of 1/2. In other words, the slope of the tangent line to the curve of f(x) at x = 4 is 1/2. Isn't that neat?
Conclusion: We Did It!
Guys, we did it! We successfully found f'(4) for the function f(x) = √(5x + 5) using the definition of the derivative. We started with a seemingly complex problem, but we broke it down into manageable steps. We understood the definition of the derivative, plugged in our values, rationalized the numerator, and finally evaluated the limit. And the result? f'(4) = 1/2.
This problem is a classic example of how to use the definition of the derivative, and mastering this technique is super important for understanding calculus. Remember, the key is to take things one step at a time and don't be afraid to get your hands dirty with some algebra. Calculus can be challenging, but it's also incredibly rewarding. Keep practicing, and you'll become a derivative-finding pro in no time!
So, next time you encounter a problem like this, remember our steps: understand the definition, plug in the values, rationalize if needed, and evaluate the limit. You've got this! And hey, if you enjoyed this walkthrough, let me know in the comments. Maybe we can tackle another calculus problem together soon. Keep those brains buzzing!