Find X Where F(x) And G(x) Have Parallel Tangent Lines

Finding Where Functions Have Parallel Tangent Lines

Hey guys! Today we're diving into a super cool calculus problem that will really test your understanding of derivatives and how they relate to the geometry of graphs. We're going to explore a scenario where we have two functions, f(x) = 3e^(2x) and g(x) = 6x³, and we need to figure out at what specific value of x their graphs will have parallel tangent lines. This is a classic problem, and it's all about understanding that the derivative of a function at a point gives us the slope of the tangent line at that point. So, if two tangent lines are parallel, it means they have the same slope. Our mission, should we choose to accept it, is to find that magical x value where the slopes of the tangent lines for f(x) and g(x) are equal. Let's get this party started!

Understanding Parallel Tangent Lines Through Derivatives

Alright, let's get down to business, guys. The core concept here is that the derivative of a function represents the instantaneous rate of change, which, when visualized graphically, is the slope of the tangent line at any given point. So, when we talk about two graphs having parallel tangent lines at a specific x value, we're essentially saying that the slopes of their respective tangent lines are identical at that x. In mathematical terms, this means that the derivative of function f at x, denoted as f'(x), must be equal to the derivative of function g at x, denoted as g'(x). So, our primary goal is to find the x value(s) that satisfy the equation f'(x) = g'(x). This is where the real problem-solving kicks in. We need to meticulously find the derivatives of both f(x) and g(x) and then set them equal to each other. This will give us a new equation involving x, which we'll then need to solve. Keep in mind that this equation might not be a simple linear one; it could involve exponential terms and polynomial terms, which might require some clever algebraic manipulation or even numerical methods to solve. But fear not, we'll walk through it step-by-step to make sure everyone's on the same page. Remember, the derivative is your best friend when dealing with slopes and rates of change in calculus. It's the key that unlocks the relationship between the algebraic form of a function and its geometric interpretation on a graph. So, let's get our hands dirty and calculate those derivatives!

Calculating the Derivatives: The First Step

Now, let's get our hands dirty and calculate the derivatives of our two functions, f(x) = 3e^(2x) and g(x) = 6x³. This is a crucial step, guys, because, as we've established, the derivatives will give us the slopes of the tangent lines. First, let's tackle f(x) = 3e^(2x). To find f'(x), we'll need to use the chain rule. The derivative of e^u is e^u * u', and in our case, u = 2x. So, the derivative of 2x with respect to x is 2. Therefore, applying the chain rule, f'(x) = 3 * e^(2x) * 2. Simplifying this, we get a clean f'(x) = 6e^(2x). Awesome! Now, let's move on to g(x) = 6x³. This one is a bit more straightforward; we'll use the power rule, which states that the derivative of ax^n is anx^(n-1). Here, a = 6 and n = 3. So, applying the power rule, g'(x) = 6 * 3 * x^(3-1). Simplifying this, we get g'(x) = 18x². Fantastic! We now have the derivatives for both functions: f'(x) = 6e^(2x) and g'(x) = 18x². These expressions represent the slopes of the tangent lines to f(x) and g(x) respectively, at any given x value. The next step is to use these derivatives to find where the slopes are equal, which will tell us where the tangent lines are parallel.

Setting the Derivatives Equal: The Core Equation

Alright, we've done the heavy lifting of calculating the derivatives. Now comes the moment of truth, guys: setting f'(x) equal to g'(x) to find the x value where the tangent lines are parallel. Remember, parallel lines have the same slope, and the derivatives give us those slopes. So, we need to solve the equation:

6e^(2x) = 18x²

This is our central equation. To simplify it a bit, we can divide both sides by 6:

e^(2x) = 3x²

Now, we have a transcendental equation. This means it's an equation that involves a mix of exponential functions and polynomial functions, and it's generally not possible to solve analytically (meaning, we can't isolate x using standard algebraic techniques alone). This is a common situation in calculus and higher mathematics. When faced with such equations, we often turn to numerical methods or graphical analysis to find approximate solutions. For this problem, since we're given multiple-choice options, we can use those options to test which value of x best satisfies this equation. Alternatively, we could use a graphing calculator or software to plot both y = e^(2x) and y = 3x² and find the x-coordinate(s) of their intersection points. The intersection points represent the x values where e^(2x) = 3x² holds true. Let's keep this equation in mind as we look at the answer choices.

Testing the Options: Finding the Solution

We've arrived at the equation e^(2x) = 3x², and as we discussed, this isn't easily solved algebraically. So, the most practical approach here, especially with multiple-choice answers, is to plug in the given values of x and see which one makes the equation as close to true as possible. Let's go through the options provided:

• A) x = -0.701 e^(2 * -0.701) ≈ e^(-1.402) ≈ 0.246 3 * (-0.701)² ≈ 3 * 0.4914 ≈ 1.474 0.246 ≠ 1.474 (Not close)

• B) x = -0.567 e^(2 * -0.567) ≈ e^(-1.134) ≈ 0.321 3 * (-0.567)² ≈ 3 * 0.3215 ≈ 0.965 0.321 ≠ 0.965 (Not close)

• C) x = -0.391 e^(2 * -0.391) ≈ e^(-0.782) ≈ 0.457 3 * (-0.391)² ≈ 3 * 0.1529 ≈ 0.459 0.457 ≈ 0.459 (Very close!)

• D) x = -0.302 e^(2 * -0.302) ≈ e^(-0.604) ≈ 0.547 3 * (-0.302)² ≈ 3 * 0.0912 ≈ 0.274 0.547 ≠ 0.274 (Not close)

• E) x = -0.258 e^(2 * -0.258) ≈ e^(-0.516) ≈ 0.596 3 * (-0.258)² ≈ 3 * 0.0666 ≈ 0.199 0.596 ≠ 0.199 (Not close)

As you can see, when we plug in x = -0.391, the values of e^(2x) and 3x² are extremely close. This indicates that x = -0.391 is the value where the graphs of f(x) and g(x) have parallel tangent lines. The slight difference is due to rounding in our calculations, but it's negligible compared to the other options.

Conclusion: The Power of Derivatives in Geometry

So there you have it, folks! By understanding that parallel tangent lines mean equal slopes, and that derivatives give us those slopes, we were able to set up an equation f'(x) = g'(x). This led us to 6e^(2x) = 18x², which simplified to e^(2x) = 3x². Because this is a transcendental equation, we couldn't solve it analytically. However, by strategically testing the provided answer choices, we found that x = -0.391 makes both sides of the equation nearly equal. This confirms that at x = -0.391, the graphs of f(x) = 3e^(2x) and g(x) = 6x³ have parallel tangent lines. This problem really highlights how calculus, particularly derivatives, provides a powerful bridge between algebraic functions and their geometric properties, like the slopes of tangent lines. Keep practicing these types of problems, guys, and you'll master the relationship between functions and their graphs in no time! It's a fundamental concept in understanding calculus.