Hey guys! Today, we're diving deep into the fascinating world of calculus, specifically focusing on the function f(x) = 1/x + ln(x). We're going to break down how to calculate its first and second derivatives step-by-step. So, grab your thinking caps, and let's get started!
Calculating the First Derivative, f'(x)
Alright, let's tackle the first derivative. When we talk about the first derivative, f'(x), we're essentially looking at the instantaneous rate of change of the function f(x). In simpler terms, it tells us how the function is changing at any given point. To find this, we need to apply some basic differentiation rules.
Our function is f(x) = 1/x + ln(x). We can rewrite 1/x as x⁻¹ to make it easier to differentiate. Now, we have:
f(x) = x⁻¹ + ln(x)
To differentiate this, we'll use the power rule and the derivative of the natural logarithm. The power rule states that if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. The derivative of ln(x) is simply 1/x. Applying these rules, we get:
f'(x) = d/dx (x⁻¹) + d/dx (ln(x))
f'(x) = -1 * x⁻² + 1/x
f'(x) = -x⁻² + 1/x
Now, let's rewrite x⁻² as 1/x² to get a cleaner expression:
f'(x) = -1/x² + 1/x
To simplify further, we can find a common denominator, which is x²:
f'(x) = -1/x² + x/x²
f'(x) = (x - 1) / x²
So, there you have it! The first derivative of f(x) = 1/x + ln(x) is f'(x) = (x - 1) / x². This expression is super important because it helps us identify critical points where the function's slope is either zero or undefined. These critical points can be potential maxima or minima, which are crucial in optimization problems. Understanding the behavior of f'(x) allows us to sketch the graph of f(x) and analyze its increasing and decreasing intervals.
Analyzing the First Derivative
The first derivative, as we've established, is a powerful tool for understanding a function's behavior. The first thing we usually look for are the critical points. These are the points where the derivative equals zero or is undefined. Setting f'(x) = 0, we find:
(x - 1) / x² = 0
This is zero when the numerator is zero, so x - 1 = 0, which means x = 1. The derivative is undefined when the denominator is zero, so x² = 0, which means x = 0. However, since the original function f(x) = 1/x + ln(x) is not defined at x = 0 (due to the 1/x and ln(x) terms), we don't consider x = 0 as a critical point in the traditional sense for this function. It's more of a point of discontinuity.
Now, we'll use a sign chart to analyze the intervals determined by our critical point x = 1 and the point of discontinuity x = 0. We'll test the intervals (0, 1) and (1, ∞) to see where f'(x) is positive or negative.
- For 0 < x < 1, let's pick x = 0.5. Then f'(0.5) = (0.5 - 1) / (0.5)² = -0.5 / 0.25 = -2, which is negative.
- For x > 1, let's pick x = 2. Then f'(2) = (2 - 1) / 2² = 1 / 4, which is positive.
This tells us that f(x) is decreasing on the interval (0, 1) and increasing on the interval (1, ∞). At x = 1, we have a local minimum. The value of the function at this point is f(1) = 1/1 + ln(1) = 1 + 0 = 1. So, we have a local minimum at the point (1, 1).
Understanding the increasing and decreasing intervals, along with the local minimum, gives us a solid foundation for sketching the graph of f(x). We know it decreases from the discontinuity at x = 0 until it reaches the minimum at (1, 1), and then it increases as x goes to infinity.
Calculating the Second Derivative, f''(x)
Next up, we're tackling the second derivative, f''(x). This tells us about the concavity of the function f(x). Is the function curving upwards (concave up) or downwards (concave down)? The second derivative is the derivative of the first derivative, so we'll be working with the expression we found earlier, f'(x) = (x - 1) / x².
To find f''(x), we need to differentiate f'(x). We'll use the quotient rule, which states that if f(x) = u(x) / v(x), then f'(x) = (v(x)u'(x) - u(x)v'(x)) / (v(x))². In our case, u(x) = x - 1 and v(x) = x².
First, let's find the derivatives of u(x) and v(x):
u'(x) = d/dx (x - 1) = 1
v'(x) = d/dx (x²) = 2x
Now, applying the quotient rule:
f''(x) = (x² * 1 - (x - 1) * 2x) / (x²)²
f''(x) = (x² - 2x² + 2x) / x⁴
f''(x) = (-x² + 2x) / x⁴
We can factor out an x from the numerator:
f''(x) = x(-x + 2) / x⁴
And then simplify by canceling an x:
f''(x) = (-x + 2) / x³
So, the second derivative of f(x) = 1/x + ln(x) is f''(x) = (2 - x) / x³. This expression is crucial for determining the concavity of the function and identifying inflection points, where the concavity changes.
Analyzing the Second Derivative
The second derivative, f''(x) = (2 - x) / x³, gives us insights into the concavity of the function f(x). To analyze it, we first need to find the points where f''(x) = 0 or is undefined. These points are potential inflection points, where the concavity of the function might change.
Setting f''(x) = 0, we have:
(2 - x) / x³ = 0
This is zero when the numerator is zero, so 2 - x = 0, which means x = 2. The second derivative is undefined when the denominator is zero, so x³ = 0, which means x = 0. Again, x = 0 is a point of discontinuity for the original function, so it's not an inflection point in the same sense.
Now, we'll create a sign chart to analyze the intervals determined by x = 0 and x = 2. We'll test the intervals (0, 2) and (2, ∞) to see where f''(x) is positive or negative.
- For 0 < x < 2, let's pick x = 1. Then f''(1) = (2 - 1) / 1³ = 1 / 1 = 1, which is positive.
- For x > 2, let's pick x = 3. Then f''(3) = (2 - 3) / 3³ = -1 / 27, which is negative.
This tells us that f(x) is concave up on the interval (0, 2) and concave down on the interval (2, ∞). At x = 2, there is an inflection point. The value of the function at this point is f(2) = 1/2 + ln(2) ≈ 0.5 + 0.693 = 1.193. So, there is an inflection point at approximately (2, 1.193).
The concavity information, along with our earlier analysis of increasing and decreasing intervals, gives us a very clear picture of the shape of f(x). It's concave up until x = 2, then it changes to concave down. This, combined with the local minimum at (1, 1), helps us sketch an accurate graph of the function.
Putting it All Together: Sketching the Graph
Okay, guys, we've done the hard work! We've calculated the first and second derivatives and analyzed their behavior. Now, we're ready to put it all together and get a good idea of what the graph of f(x) = 1/x + ln(x) looks like. Here’s a recap of what we know:
- Domain: The function is defined for x > 0 due to the ln(x) term.
- First Derivative: f'(x) = (x - 1) / x². Critical point at x = 1. Decreasing on (0, 1), increasing on (1, ∞). Local minimum at (1, 1).
- Second Derivative: f''(x) = (2 - x) / x³. Inflection point at x = 2. Concave up on (0, 2), concave down on (2, ∞).
With this information, we can sketch the graph. We know there's a vertical asymptote at x = 0. The function decreases from x = 0 until it reaches the local minimum at (1, 1). After that, it increases towards infinity. The concavity changes at x = 2, so the graph curves upwards until (2, 1.193), and then it starts curving downwards.
The graph will have a smooth, continuous shape (except at x = 0) and will clearly show the local minimum and the inflection point. Sketching the graph is an excellent way to visualize the behavior of the function and confirm our calculations.
Conclusion
So there you have it! We've successfully calculated the first and second derivatives of f(x) = 1/x + ln(x) and used them to analyze the function's behavior. We found critical points, intervals of increase and decrease, concavity, and inflection points. This detailed analysis allows us to understand and sketch the graph of the function accurately.
Remember, derivatives are powerful tools in calculus, and mastering them opens doors to understanding more complex functions and solving real-world problems. Keep practicing, and you'll become a calculus whiz in no time! If you guys have any questions, feel free to ask. Happy calculating!
