Derivative Of H(x) To The Left Of 3: A Calculus Problem

Let's dive into this interesting calculus problem where we need to find the derivative of a piecewise function at a specific point. Specifically, we're looking at the function $h(x)$ which is defined differently for values of $x$ less than or equal to 3, and for values of $x$ greater than 3. Our mission, should we choose to accept it, is to determine the derivative of $h(x)$ to the left of 3. This means we're interested in how the function is changing as $x$ approaches 3 from values smaller than 3. So, buckle up, math enthusiasts, because we're about to embark on a derivative-finding adventure!

Understanding the Piecewise Function

Before we jump into the derivatives, let's make sure we're all on the same page about what this piecewise function, $h(x)$, actually looks like. A piecewise function, as the name suggests, is a function that's defined in "pieces." In our case, $h(x)$ has two pieces:

1. When $x$ is less than or equal to 3 ($x e 3$), $h(x)$ is defined as $x - 1$. This is a simple linear function, a straight line with a slope of 1 and a y-intercept of -1. Think of it as a friendly line gently sloping upwards as you move from left to right on the graph.
2. When $x$ is greater than 3 ($x > 3$), $h(x)$ takes on a different persona and is defined as $2x - 4$. This is another linear function, but this time it's a bit steeper, with a slope of 2 and a y-intercept of -4. This line climbs more rapidly than our first one.

The key takeaway here is that the function's behavior changes at $x = 3$. It's like a chameleon changing colors depending on its surroundings. So, to find the derivative to the left of 3, we need to focus on the piece of the function that's active when $x$ is approaching 3 from the left – that's the $x - 1$ part.

Finding the Derivative to the Left of 3

Okay, now for the main event: finding the derivative. Remember, the derivative of a function at a point tells us the instantaneous rate of change of the function at that point. In simpler terms, it's the slope of the line tangent to the function's graph at that point. Since we're interested in the derivative to the left of 3, we'll only consider the piece of the function defined for $x e 3$, which is $h(x) = x - 1$.

The beauty of this situation is that $x - 1$ is a linear function. And a fundamental concept in calculus is that the derivative of a linear function is simply its slope. The slope of the line $y = mx + b$ is always $m$. In our case, $h(x) = x - 1$ can be written as $h(x) = 1x - 1$, so the slope (and therefore the derivative) is 1.

To be a little more formal, we can use the power rule for differentiation. The power rule states that the derivative of $x^n$ is $nx^{n-1}$. Applying this to $h(x) = x - 1$:

• The derivative of $x$ (which is $x^1$) is $1 * x^(1-1) = 1 * x^0 = 1$.
• The derivative of the constant -1 is 0 (constants always have a derivative of 0).

Therefore, the derivative of $h(x) = x - 1$ is simply 1. This means that no matter what value of $x$ we choose (as long as it's less than or equal to 3), the function is changing at a constant rate of 1. The graph of the function is a straight line, so its slope is constant.

The Answer and Why It Matters

So, the derivative of $h(x)$ to the left of 3 is 1. Looking at our options, that corresponds to answer choice (C). But beyond just getting the right answer, what does this tell us? It tells us that as $x$ approaches 3 from the left, the function $h(x)$ is increasing at a rate of 1. For every small increase in $x$, $h(x)$ increases by the same amount. This constant rate of change is a hallmark of linear functions.

This problem highlights a crucial concept in calculus: how to deal with piecewise functions and derivatives. When a function is defined differently over different intervals, we need to be careful about which piece of the function we're considering when we calculate the derivative. In this case, we focused on the piece of the function that was relevant to the left of 3, and that led us to the correct answer.

Additional Considerations: Differentiability at x = 3

While we've found the derivative to the left of 3, it's worth pondering what happens at $x = 3$. To be differentiable at a point, a function needs to be both continuous and have matching left-hand and right-hand derivatives at that point. Let's investigate:

• Continuity: Is $h(x)$ continuous at $x = 3$? For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function's value at that point must all be equal.
  • The left-hand limit as $x$ approaches 3 is $3 - 1 = 2$.
  • The right-hand limit as $x$ approaches 3 is $2(3) - 4 = 2$.
  • The function's value at $x = 3$ is $3 - 1 = 2$. So, $h(x)$ is indeed continuous at $x = 3$.
• Matching Derivatives: We found the derivative to the left of 3 to be 1. What's the derivative to the right of 3? For $x > 3$, $h(x) = 2x - 4$, which has a derivative of 2 (its slope). Since the left-hand derivative (1) doesn't match the right-hand derivative (2), $h(x)$ is not differentiable at $x = 3$.

This is an important observation: a function can be continuous at a point but still not be differentiable there. This often happens at "corners" or "sharp turns" in the graph of the function, which is exactly what we have in this case at $x = 3$.

Conclusion

So, there you have it! We've successfully navigated the world of piecewise functions and derivatives to find that the derivative of $h(x)$ to the left of 3 is 1. We also explored the concept of differentiability and discovered that while $h(x)$ is continuous at $x = 3$, it's not differentiable there due to the mismatch in left-hand and right-hand derivatives. This problem serves as a great reminder of the nuances of calculus and the importance of understanding the definitions and concepts thoroughly. Keep practicing, and you'll be a derivative-detecting pro in no time!