Problem 18 - Precalculus: Mathematical for Calculus

Question

Find the derivative of the function at the given number. $$G(x)=1+2 \sqrt{x} \text { at } 4$$

Step-by-Step Solution

Answer

The derivative of the function at 4 is \( \frac{1}{2} \).

Step 1: Identify the Function and the Point

We are given the function \( G(x) = 1 + 2 \sqrt{x} \) and need to find the derivative at \( x = 4 \). This means we are looking for \( G'(4) \).

Step 2: Differentiate the Function

To differentiate \( G(x) = 1 + 2 \sqrt{x} \), recall that \( \sqrt{x} \) can be rewritten as \( x^{1/2} \). The derivative of \( x^{1/2} \) is \( \frac{1}{2}x^{-1/2} \). Therefore, the derivative of \( 2\sqrt{x} = 2x^{1/2} \) is \( 2 \times \frac{1}{2}x^{-1/2} = x^{-1/2} \). The derivative of a constant \( 1 \) is \( 0 \). Thus, \( G'(x) = 0 + x^{-1/2} = \frac{1}{\sqrt{x}} \).

Step 3: Evaluate the Derivative at the Given Point

Substitute \( x = 4 \) into the derivative \( G'(x) = \frac{1}{\sqrt{x}} \). This gives us \( G'(4) = \frac{1}{\sqrt{4}} = \frac{1}{2} \).

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus, often representing how a function changes at any given point. Imagine driving a car and looking at the speedometer. Derivatives are like the speedometer readings, showing the instant speed or the rate of change of position. To find a derivative, you need to differentiate the given function. This involves using rules and techniques to calculate how the output value (function result) changes in response to a small change in the input value (usually denoted by \(x\)).

In mathematical terms, given a function \(f(x)\), its derivative \(f'(x)\) gives us the slope of the tangent to the curve at any point \(x\). The steps often involve understanding basic derivatives of functions like powers of \(x\), constants, and applying rules.

• The derivative of a constant is always \(0\).
• The power rule states that \( \frac{d}{dx} x^n = nx^{n-1} \).
• Chain and product rules might be necessary for more complex expressions.

Derivatives help us understand rates of change and solve many real-world problems like motion modeling and finding optimal solutions.

Function Evaluation

Function evaluation is the process of replacing the variable in a function with a specific number and calculating the resulting value. Think of it like testing your favorite recipe. Once you have all the ingredients (numbers) ready and included, the outcome is your delicious dish (the function's value).

In the context of derivatives, once we have differentiated the function, we often need to evaluate it at a specific point. This means substituting the given value into the derivative, just like substituting an ingredient measurement into a recipe.
To evaluate a function:

• Replace every instance of the variable with the specified value.
• Simplify the expression to get the result.

In our example, after finding the derivative \(G'(x) = \frac{1}{\sqrt{x}}\), evaluating it at \(x = 4\) means calculating \(G'(4)\). The function evaluates to \(\frac{1}{\sqrt{4}} = \frac{1}{2}\), giving us the rate of change at that specific point.

Square Root Function

The square root function is an important and frequently encountered function in mathematics, represented as \( \sqrt{x} \). It is the inverse of squaring a number, essentially asking "what number multiplied by itself gives me \(x\)?".

Some important characteristics of the square root function include:

• It's only defined for non-negative numbers if real numbers are considered.
• The graph of \(y = \sqrt{x}\) is a smooth curve starting at the origin (0,0) and increasing steadily.
• Its derivative is helpful when analyzing the behavior or rate of change of the function.

When differentiating a square root function, it's often helpful to express it in an exponent form \(x^{1/2}\). This aids in applying the power rule for differentiation. Thus, the derivative of \(\sqrt{x}\) is \( \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \). Understanding this transformation is key to handling derivatives involving square roots and solving related problems.