Problem 23 - Single Variable Calculus

Question

Find the linearization of \(f(x)=\sqrt{x+3}\) at \(a=1\), and use it to approximate the numbers \(\sqrt{3.9}\) and \(\sqrt{4.1}\). Plot the graphs of \(f\) and \(L\) on the same set of axes.

Step-by-Step Solution

Answer

The linearization of the function \(f(x)=\sqrt{x+3}\) at \(a=1\) is given by \(L(x) = 2 + \frac{1}{4}(x-1)\). Using this linearization, we can approximate \(\sqrt{3.9}\) as \(1.975\) and \(\sqrt{4.1}\) as \(2.025\). When plotting both the function \(f(x)\) and its linearization \(L(x)\) on the same graph, the straight line of \(L(x)\) is tangent to the curve of \(f(x)\) at the point \(a=1\).

Step 1: Find f(a) at a = 1

To find \(f(a)\), we simply evaluate the function \(f(x)\) at \(x = a = 1\). \[f(1) = \sqrt{1+3} = \sqrt{4} = 2\]

Step 2: Find f'(x) and f'(a)

Now, we will find the derivative of \(f(x)\) with respect to x. Given the function \(f(x) = \sqrt{x + 3}\), we will apply the chain rule to find its derivative. The chain rule states that \(\frac{d}{dx}[g(h(x))] = g'(h(x))h'(x)\). In this case, we have \(g(u) = \sqrt{u}\) and \(h(x) = x + 3\). The derivatives of these two functions are: \(g'(u) = \frac{1}{2\sqrt{u}} \hspace{10mm} h'(x) = 1\) Now we apply the chain rule: \[ f'(x) = g'(h(x))h'(x) = \frac{1}{2\sqrt{x+3}} \cdot 1 = \frac{1}{2\sqrt{x+3}} \] Now we will find \(f'(a)\) at \(a=1\): \[ f'(1) = \frac{1}{2\sqrt{1+3}} = \frac{1}{4} \]

Step 3: Find the linearization L(x)

Now that we have \(f(a)\) and \(f'(a)\), we can find the linearization \(L(x)\) using the following formula: \[L(x) = f(a) + f'(a)(x - a) = 2 + \frac{1}{4}(x - 1)\]

Step 4: Approximate the values

We will now use the linearization to approximate the values \(\sqrt{3.9}\) and \(\sqrt{4.1}\): \[L(0.9) \approx \sqrt{3.9} = 2 + \frac{1}{4}(0.9 - 1) = 2 - \frac{1}{4}(0.1) = 2 - 0.025 = 1.975\] \[L(1.1) \approx \sqrt{4.1} = 2 + \frac{1}{4}(1.1 - 1) = 2 + \frac{1}{4}(0.1) = 2 + 0.025 = 2.025\]

Step 5: Plot f(x) and L(x)

To plot \(f(x) = \sqrt{x + 3}\) and \(L(x) = 2 + \frac{1}{4}(x - 1)\), the domain for both functions must be determined. Given that the exercise is focused on the interval around the point \(a = 1\), a suitable plot interval would be from \(0\) to \(2\). On a graph, plot the function \(f(x) = \sqrt{x + 3}\) and its linearization \(L(x) = 2 + \frac{1}{4}(x - 1)\) from \(0\) to \(2\). The non-linear curve represents the function \(f(x)\), and the straight line represents the linearization \(L(x)\), which is tangent to the curve of the function at the point \(a = 1\).

Key Concepts

Derivative

Understanding derivatives is a key part of studying calculus and analyzing functions. Derivatives represent the rate at which a function changes at any given point. In simpler terms, it shows how much a function's output changes in response to a small change in its input.
For the function \(f(x) = \sqrt{x+3}\), the derivative \(f'(x)\) measures the slope of the tangent line to the graph of \(f(x)\) at any point \(x\). By applying the chain rule, a powerful technique for finding derivatives of composite functions, we determined \(f'(x) = \frac{1}{2\sqrt{x+3}}\).
This derivative formula helps us understand how quickly or slowly the function \(f(x)\) is rising or falling. Evaluating this derivative at the point \(a = 1\) gives us the specific slope \(f'(1) = \frac{1}{4}\), which is crucial for finding a tangent line to the curve at this point.

Function Approximation

One of the most useful applications of derivatives is in approximating complicated functions with simpler ones. In this context, we use the concept of linearization to approximate a function near a given point.
Linearization involves creating a linear function \(L(x)\) that closely matches the original function \(f(x)\) near \(x = a\). Here, we've derived \(L(x) = 2 + \frac{1}{4}(x - 1)\). This equation produces a straight line that closely resembles the behavior of \(f(x)\) near \(a = 1\).
This linear approximation is especially handy when you need to quickly estimate the function for values close to \(a\). We utilized \(L(x)\) to approximate \(\sqrt{3.9}\) as \(1.975\) and \(\sqrt{4.1}\) as \(2.025\), showcasing how linearization provides a practical solution for estimating function values.

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of composite functions. A composite function is a function that contains another function, much like nesting your favorite toy inside another toy box.
In our exercise, the chain rule was crucial in differentiating \(f(x) = \sqrt{x + 3}\). We identified \(g(u) = \sqrt{u}\) and \(h(x) = x + 3\) to apply the rule effectively. According to the chain rule, we take the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Mathematically, it gave us:

• \(g'(u) = \frac{1}{2\sqrt{u}}\)
• \(h'(x) = 1\)

Combining these using the chain rule gives \(f'(x) = \frac{1}{2\sqrt{x + 3}}\). This step-by-step approach simplifies the complexity of composite differentiation, empowering us to derive the function's rate of change efficiently.

Plotting Graphs

Visualizing functions and their approximations enhances our understanding, as graphs provide a picture of how a function behaves. When we plot the function \(f(x) = \sqrt{x+3}\) alongside its linearization \(L(x)\), we gain valuable insights into both.
Choosing an appropriate domain, such as from \(0\) to \(2\), is vital since it focuses on the area around the specific point of interest, \(a = 1\). This visualization allows us to see how well the linear approximation matches the actual function around that point. The graph of \(f(x)\) forms a curve that rises steadily, while the straight line \(L(x)\) represents the linearization.
By plotting both together, we can visually assess where the linear approximation works well and where it might begin to differ from the true function. Such graphical analysis is invaluable in mathematics, making complex information intuitive and easy to grasp.