Horizontal Tangent: Find X For F(x) = X³ - 6x² + 9x + 2
Hey guys! Let's dive into a super common calculus problem: finding where a function has a horizontal tangent. Specifically, we're going to figure out at what x-values the function f(x) = x³ - 6x² + 9x + 2 has a slope of zero. This is a classic application of derivatives, so buckle up, and let’s get started!
Understanding Horizontal Tangents
So, what exactly is a horizontal tangent? Think about it visually. A tangent line is a line that touches a curve at only one point (locally, at least). When that tangent line is perfectly horizontal, it means the function isn't increasing or decreasing at that specific point. In calculus terms, the slope of the tangent line is given by the derivative of the function, f'(x). Therefore, a horizontal tangent occurs where the derivative equals zero: f'(x) = 0. This is a crucial concept. Whenever you hear "horizontal tangent," your brain should immediately think, "derivative equals zero!"
Why is this important? Because horizontal tangents often indicate local maxima (peaks) or local minima (valleys) of the function. They're critical points that help us understand the function's behavior. Finding these points allows us to sketch the graph of the function, determine its range, and solve optimization problems. Basically, knowing where the horizontal tangents are gives you a ton of information about the function itself. Also horizontal tangent lines can occur at saddle points.
Finding the Derivative
Okay, let's get our hands dirty with the actual function: f(x) = x³ - 6x² + 9x + 2. The first step is to find its derivative, f'(x). Remember the power rule? It states that if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. We'll apply this rule to each term in our function. The derivative of x³ is 3x². The derivative of -6x² is -12x. The derivative of 9x is 9. And the derivative of the constant 2 is 0. Putting it all together, we get: f'(x) = 3x² - 12x + 9. This quadratic expression represents the slope of the tangent line to the original function at any given value of x.
Derivatives are fundamental in calculus. They tell us the instantaneous rate of change of a function. In simpler terms, they tell us how much a function is changing at any given point. For example, in physics, if f(x) represents the position of an object at time x, then f'(x) represents the object's velocity at time x. Understanding and being able to compute derivatives quickly is essential for mastering calculus and its applications.
Setting the Derivative to Zero
Now comes the key step: we need to find the values of x where f'(x) = 0. In other words, we need to solve the equation 3x² - 12x + 9 = 0. This is a quadratic equation, and we have a few options for solving it: factoring, using the quadratic formula, or completing the square. Factoring is often the quickest method if it's possible.
Looking at our equation, 3x² - 12x + 9 = 0, we can simplify it by dividing both sides by 3: x² - 4x + 3 = 0. Now, can we factor this? We're looking for two numbers that multiply to 3 and add up to -4. Those numbers are -1 and -3. So, we can factor the quadratic as (x - 1)(x - 3) = 0. To find the solutions for x, we set each factor equal to zero:
- x - 1 = 0 => x = 1
- x - 3 = 0 => x = 3
Therefore, the solutions are x = 1 and x = 3. These are the x-values where the function f(x) = x³ - 6x² + 9x + 2 has a horizontal tangent.
Verifying the Solutions
It's always a good idea to double-check your work, especially in calculus! We can verify our solutions by plugging them back into the derivative f'(x) and making sure we get zero. Let's try x = 1: f'(1) = 3(1)² - 12(1) + 9 = 3 - 12 + 9 = 0. Great! Now let's try x = 3: f'(3) = 3(3)² - 12(3) + 9 = 27 - 36 + 9 = 0. Excellent! Both values of x make the derivative equal to zero, confirming that we have found the correct locations of the horizontal tangents.
Another way to visualize this is to think about the graph of the function. At x = 1 and x = 3, the function momentarily stops increasing or decreasing, creating a flat spot on the curve. If you were to zoom in close enough at these points, the tangent line would appear perfectly horizontal.
The Answer
So, to answer the original question: The function f(x) = x³ - 6x² + 9x + 2 has a horizontal tangent at x = 1 and x = 3. Therefore, the correct answer is x=1 and x=3.
Why This Matters
Understanding horizontal tangents isn't just about solving textbook problems. It's a fundamental concept that has applications in many fields, including:
- Optimization: Finding the maximum or minimum values of a function is crucial in engineering, economics, and other areas. Horizontal tangents help us identify these extreme values.
- Curve Sketching: Knowing where a function has horizontal tangents allows us to accurately sketch its graph and understand its behavior.
- Physics: As mentioned earlier, derivatives can represent velocity and acceleration. Finding horizontal tangents can help us determine when an object changes direction or reaches its maximum height.
- Machine Learning: Gradient descent, a common optimization algorithm in machine learning, relies on finding points where the derivative is close to zero to minimize a loss function.
Practice Problems
To solidify your understanding, try these practice problems:
- Find the x-values where the function g(x) = 2x³ + 3x² - 12x + 5 has a horizontal tangent.
- Find the x-values where the function h(x) = x⁴ - 8x² + 16 has a horizontal tangent.
Remember to follow the same steps we used in the example: find the derivative, set it equal to zero, and solve for x. Good luck, and happy calculating!
Conclusion
Finding horizontal tangents is a core skill in calculus. By understanding the relationship between the derivative and the slope of a tangent line, you can solve a wide range of problems and gain valuable insights into the behavior of functions. So keep practicing, and don't be afraid to ask questions. You got this! And remember, when you see "horizontal tangent," think f'(x) = 0!