Linearize F(x) = 2x^3 - 3x^2 At X = 2: Exact Answer

When we talk about linearization, we're essentially trying to approximate a complex function with a simpler one – a straight line – around a specific point. This is incredibly useful in calculus and various scientific fields because linear functions are much easier to work with than higher-order polynomials or other complex curves. The process of linearization allows us to understand the local behavior of a function. Instead of dealing with the full complexity of $f(x)=2 x^3-3 x^2$, we can find a line that closely matches its value and slope at $x=2$. This approximation is particularly valuable when we're interested in what happens near $x=2$. Think of it like zooming in on a curve with a magnifying glass; from a close enough perspective, even a curve can look remarkably like a straight line. The mathematical tool we use for this is the tangent line. The tangent line to a function at a given point is the unique line that touches the function at that point and has the same slope as the function at that point. This tangent line serves as our linearization. To find this tangent line, we need two key pieces of information: the value of the function at the point of interest and the slope of the function at that point. The value of the function at $x=2$ gives us a point the line must pass through, and the slope at $x=2$ tells us the direction of our approximating line. The formula for the linearization $L(x)$ of a function $f(x)$ at a point $x=a$ is given by $L(x) = f(a) + f'(a)(x-a)$. Here, $f(a)$ is the function's value at $x=a$, and $f'(a)$ is the derivative of the function evaluated at $x=a$, which represents the slope of the tangent line at that point. Our goal is to find this exact linearization for $f(x)=2 x^3-3 x^2$ at the specific point $x=2$. This means we need to calculate $f(2)$ and $f'(2)$ with precision, ensuring our final answer is exact, not an approximation of an approximation.

Calculating $f(a)$ and $f'(a)$ for our Function

The first step in finding the linearization of $f(x)=2 x^3-3 x^2$ at $x=2$ involves calculating the value of the function at this point. So, we need to find $f(2)$. We substitute $x=2$ into our function: $f(2) = 2(2)^3 - 3(2)^2$. Let's break this down: $(2)^3$ is $2 imes 2 imes 2 = 8$, and $(2)^2$ is $2 imes 2 = 4$. So, $f(2) = 2(8) - 3(4)$. This simplifies to $f(2) = 16 - 12$, which gives us $f(2) = 4$. This tells us that the point $(2, 4)$ lies on the graph of our function $f(x)$. This point will be crucial for our linearization, as the tangent line must pass through it. Now, we need to determine the slope of the function at $x=2$. The slope of a function at any given point is found by its derivative. So, we need to find the derivative of $f(x)$, denoted as $f'(x)$. Our function is $f(x) = 2x^3 - 3x^2$. Using the power rule for differentiation, which states that the derivative of $x^n$ is $nx^{n-1}$, we can find the derivative of each term. For the first term, $2x^3$, the derivative is $2 imes 3x^{3-1} = 6x^2$. For the second term, $-3x^2$, the derivative is $-3 imes 2x^{2-1} = -6x^1 = -6x$. Therefore, the derivative of our function is $f'(x) = 6x^2 - 6x$. With the derivative $f'(x)$ in hand, we can now find the slope of the tangent line at our specific point $x=2$. We do this by evaluating $f'(2)$: $f'(2) = 6(2)^2 - 6(2)$. Let's compute this: $(2)^2 = 4$, so $f'(2) = 6(4) - 6(2)$. This becomes $f'(2) = 24 - 12$, which results in $f'(2) = 12$. This means that at the point $x=2$, the function $f(x)$ is increasing at a rate of 12 units vertically for every 1 unit horizontally. This value, $f'(2)=12$, is the slope of our linearization.

Constructing the Linearization Formula

Now that we have all the necessary components, we can construct the linearization, $L(x)$, of $f(x)=2 x^3-3 x^2$ at $x=2$. The general formula for linearization at a point $x=a$ is $L(x) = f(a) + f'(a)(x-a)$. In our case, $a=2$. We have already calculated $f(2) = 4$ and $f'(2) = 12$. Plugging these values into the linearization formula, we get: $L(x) = 4 + 12(x-2)$. This is the equation of the tangent line to $f(x)$ at $x=2$. To express this in a more standard linear form ($y = mx + b$), we can distribute the 12 and combine the constant terms: $L(x) = 4 + 12x - 12 imes 2$. This simplifies to $L(x) = 4 + 12x - 24$. Combining the constant terms, we get $L(x) = 12x - 20$. This is the exact linearization of $f(x)=2 x^3-3 x^2$ at $x=2$. This linear function, $L(x) = 12x - 20$, will provide a good approximation of $f(x)$ for values of $x$ that are close to 2. For example, if we wanted to approximate $f(2.01)$, we could use $L(2.01) = 12(2.01) - 20 = 24.12 - 20 = 4.12$. Let's check the actual value of $f(2.01)$: $f(2.01) = 2(2.01)^3 - 3(2.01)^2$. While calculating this by hand might be tedious, using a calculator, we find that $f(2.01) imes 2.01^3 imes 2 - 3 imes 2.01^2 imes ext{approx} 8.12 - 3 imes 4.04 imes ext{approx} 8.12 - 12.12 imes ext{approx} 16.24 - 12.12 imes ext{approx} 4.12$. This shows how close the linearization is to the actual function value near the point of tangency. The power of linearization lies in its ability to simplify complex calculations and provide insightful approximations in a local region of a function.

Understanding the Significance of Linearization

The linearization of a function at a point is a fundamental concept in calculus with wide-ranging applications. It's not just about finding a line; it's about understanding how a function behaves locally. The tangent line, which serves as our linearization, captures the instantaneous rate of change of the function at that specific point. This rate of change is precisely what the derivative measures. Therefore, the linearization $L(x) = f(a) + f'(a)(x-a)$ can be interpreted as a first-order approximation of $f(x)$ near $x=a$. The term $f(a)$ represents the value of the function at the point, ensuring our approximation starts from the correct height. The term $f'(a)(x-a)$ represents the change in the function's value as we move from $a$ to $x$, based on the slope $f'(a)$. The smaller the distance $(x-a)$, the more accurate this linear approximation tends to be. For our specific function, $f(x)=2 x^3-3 x^2$, and the point $x=2$, we found the linearization $L(x) = 12x - 20$. This means that for values of $x$ very close to 2, the value of $2 x^3-3 x^2$ can be closely approximated by $12x - 20$. This is incredibly useful in physics and engineering, for instance, where complex models might be simplified using linearization to make calculations feasible. Imagine trying to predict the motion of an object under a complicated force field; linearization can help simplify the equations of motion in small time intervals or around specific equilibrium points. In numerical analysis, linearization is a key step in algorithms used to solve equations or optimize functions. For example, Newton's method for finding roots of a function relies heavily on linearization. At each step, it approximates the function with its tangent line and finds where that line crosses the x-axis, providing a better estimate for the root. The process of linearization also helps build intuition about function behavior. By looking at the tangent line, we can immediately tell if the function is increasing or decreasing at that point (based on the sign of $f'(a)$) and how rapidly (based on the magnitude of $f'(a)$). In summary, the exact linearization $L(x) = 12x - 20$ for $f(x)=2 x^3-3 x^2$ at $x=2$ is more than just an equation; it's a powerful tool that simplifies complex analysis by focusing on the local linear behavior of a function. It highlights the fundamental relationship between a function, its derivative, and the tangent line that approximates it in the vicinity of a point.

Conclusion

We have successfully determined the exact linearization of the function $f(x)=2 x^3-3 x^2$ at the point $x=2$. By following the standard procedure of finding the function's value and its derivative's value at the specified point, we constructed the tangent line equation. The function value at $x=2$ is $f(2) = 4$, and the derivative at $x=2$ is $f'(2) = 12$. Using the linearization formula $L(x) = f(a) + f'(a)(x-a)$, we arrived at the exact linearization $L(x) = 4 + 12(x-2)$, which simplifies to $L(x) = 12x - 20$. This linear equation provides an accurate approximation of $f(x)$ for values of $x$ near 2, illustrating the power of calculus in simplifying complex functions locally. Understanding linearization is crucial for grasping more advanced calculus concepts and their applications in various scientific and engineering disciplines.

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