Problem 39 - Calculus Single Variable

Question

Calculate the linearization \(L(x)=f(c)+\) \(f^{\prime}(c), \cdot(x-c)\) for the given function \(f\) at the given value \(c\) $$ f(x)=(25 / 9)^{x}, c=1 / 2 $$

Step-by-Step Solution

Answer

The linearization of the function at \(c=\frac{1}{2}\) is \(L(x) = \frac{5}{3} + \frac{5}{3} \ln\left(\frac{25}{9}\right) \cdot (x-\frac{1}{2})\).

Step 1: Determine the Function Value

Compute the function value at the point \(c\). For \(f(x) = \left(\frac{25}{9}\right)^x\) and \(c = \frac{1}{2}\), calculate \(f\left(\frac{1}{2}\right)\): \[f\left(\frac{1}{2}\right) = \left(\frac{25}{9}\right)^{\frac{1}{2}} = \sqrt{\frac{25}{9}} = \frac{5}{3}\]

Step 2: Find the Derivative

Find the derivative of \(f(x) = \left(\frac{25}{9}\right)^x\). Using the natural logarithm, \(f'(x) = \left(\frac{25}{9}\right)^x \ln\left(\frac{25}{9}\right)\).

Step 3: Evaluate the Derivative at c

Substitute \(c = \frac{1}{2}\) into the derivative to find \(f'\left(\frac{1}{2}\right)\): \[f'\left(\frac{1}{2}\right) = \left(\frac{25}{9}\right)^{\frac{1}{2}} \ln\left(\frac{25}{9}\right) = \frac{5}{3} \cdot \ln\left(\frac{25}{9}\right)\]

Step 4: Substitute into the Linearization Formula

Use the linearization formula \(L(x) = f(c) + f'(c) \cdot (x-c)\). Here, \(f(c) = \frac{5}{3}\), \(f'(c) = \frac{5}{3} \ln\left(\frac{25}{9}\right)\), and \(c = \frac{1}{2}\), so \[L(x) = \frac{5}{3} + \frac{5}{3} \ln\left(\frac{25}{9}\right) \cdot \left(x - \frac{1}{2}\right)\]

Key Concepts

Derivatives

To truly comprehend the concept of linearization, it's pivotal to grasp what a derivative is. The derivative of a function fundamentally represents the rate at which that function is changing at any given point. In more straightforward terms, it tells us how "steep" the curve is at a point. Let's see how this applies to our exercise.

For the function \(f(x) = \left(\frac{25}{9}\right)^x\), finding the derivative involves recognizing that the base \(\frac{25}{9}\) is a constant raised to a power. When you have functions of this form, you apply the derivative of exponential functions which is \(f'(x) = b^x \ln(b)\), where \(b\) is the base of the exponential. Here, \(b = \frac{25}{9}\).

So, for our function, the derivative is \(f'(x) = \left(\frac{25}{9}\right)^x \ln\left(\frac{25}{9}\right)\). Notice how the natural logarithm, denoted as \(\ln\), naturally appears when differentiating exponential functions with bases other than \(e\), which we'll discuss next.

Exponential Functions

Exponential functions are a cornerstone of mathematics, often appearing in contexts ranging from biology to finance. If you're working with an exponential function, it typically looks like \(f(x) = a^x\) where \(a\) is a positive real number. They have unique properties, including growth rates that intensify rapidly.

In our problem, you have the function \(f(x) = \left(\frac{25}{9}\right)^x\). Here, \(\frac{25}{9}\) is the base of the exponential function, suggesting that this function describes a kind of exponential growth, except with a base different from the natural base \(e\).

This base transforms the expression as it philosophically nudges up or down by multiplying the preceding value at every increment of \(x\). This system renders exponential functions remarkably useful in modeling systems where growth is proportional to the current size, like compound interest or populations.

Natural Logarithm

The natural logarithm, often written as \(\ln\), is an inverse function of the exponential function with base \(e\). Understanding \(\ln\) is critical, particularly when differentiating exponential functions not base \(e\). It provides insight into how these functions grow.

The natural logarithm \(\ln\) helps mediate the transition between rates of exponential growth and simpler arithmetic growth. When you're calculating derivatives of functions like \( f(x) = \left(\frac{25}{9}\right)^x \), using \(\ln\) is crucial. It’s the tool to extract a "rate of growth"—the rate here being the coefficient of \(x\) in the exponential.

In our exercise, this manifested as \(f'(x) = \left(\frac{25}{9}\right)^x \ln\left(\frac{25}{9}\right)\). By applying \(\ln\left(\frac{25}{9}\right)\), we're evaluating how quickly \(\left(\frac{25}{9}\right)^x\) itself expands as \(x\) increases—an expansion rate pivotal in understanding not just the shape of exponential functions, but in observing their growth through derivatives.