Problem 27 - Calculus for Biology and Medicine

Question

Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=(1+x)^{-n}\) at \(a=0\). (Assume that \(n\) is a positive integer.)

Step-by-Step Solution

Answer

The linear approximation is \(L(x) = 1 - nx\).

Step 1: Evaluate the Function at a

We need to find the value of the function at the point where we are approximating, which is \(a = 0\). So, compute \(f(0) = (1+0)^{-n} = 1^{-n} = 1\).

Step 2: Find the Derivative of the Function

Calculate the derivative of \(f(x) = (1+x)^{-n}\) with respect to \(x\). Use the power rule for derivatives, which gives \(f^{\prime}(x) = -n(1+x)^{-n-1}\).

Step 3: Evaluate the Derivative at a

Now calculate the derivative at \(a = 0\): \(f^{\prime}(0) = -n(1+0)^{-n-1} = -n(1)^{-n-1} = -n\).

Step 4: Write the Linear Approximation Formula

Substitute \(f(0)\) and \(f^{\prime}(0)\) into the linear approximation formula: \(L(x) = f(0) + f^{\prime}(0)(x-a) = 1 - nx\), since \(a = 0\).

Key Concepts

Understanding Derivatives

A derivative measures how a function changes as its input changes. Imagine you're driving a car; the speed of the car at a particular instant is like the derivative of your journey's distance with respect to time. When we talk about the derivative of a function, we mean how the function's output changes concerning a small change in its input. This concept is foundational in calculus because it gives us a way to study functions' behavior at any particular point.

To find a derivative, we often use specific rules tailored to different types of functions. In our exercise, we calculated the derivative of the function \(f(x) = (1+x)^{-n}\) using these rules. Derivatives are not just mathematical tools; they also help us understand and predict real-world phenomena. Anytime you see a function, think about its derivative as a way to explore its nuances and dynamics.

Applying the Power Rule

The power rule is a handy tool for finding derivatives quickly and easily. It's particularly useful when working with polynomials or functions expressed as exponents.
For a function of the form \(x^n\), its derivative is computed as \(nx^{n-1}\). This formula emerges from applying the power rule, which simplifies derivative calculations significantly.In our exercise, we used the power rule to find the derivative of \(f(x) = (1+x)^{-n}\). We adjusted the base function by rewriting it as an exponent, helping us calculate that \(f^{\prime}(x) = -n(1+x)^{-n-1}\). This example illustrates how the power rule provides a straightforward approach to handling derivatives in problems like these, making calculus more approachable overall.

Linear Approximation Formula

The linear approximation formula is a valuable technique used to estimate the value of a function near a given point. When we cannot easily calculate a function's exact value, linear approximation allows us to use a line—specifically, a tangent line—to approximate values near the point of interest.

The formula is: \[L(x) = f(a) + f^{\prime}(a)(x-a)\]. Here, \(f(a)\) is the function's value at a specific point \(a\), and \(f^{\prime}(a)\) is the derivative at that point. Essentially, you're forming a straight line based on how the function behaves around \(a\). Using this formula in our exercise, we determined that the linear approximation of \(f(x) = (1+x)^{-n}\) at \(a=0\) is \(L(x) = 1 - nx\). This means that for values of \(x\) close to zero, \(f(x)\) can be approximated effectively with this linear expression, once more highlighting the practical use of derivatives and approximations in calculus.