Problem 19 - Calculus Single Variable

Question

A function \(f\) and a point \(c\) are given. Calculate \(f^{\prime}(c)\). $$ f(x)=3 x^{2}+2 / x \quad c=-2 $$

Step-by-Step Solution

Answer

The derivative at \( c = -2 \) is \( -12.5 \).

Step 1: Understand the Problem

You need to find the derivative of the given function, \( f(x) = 3x^2 + \frac{2}{x} \), and evaluate it at \( c = -2 \). This involves finding \( f'(x) \), the derivative of \( f \), and then substituting \( c = -2 \) into this derivative.

Step 2: Find the Derivative of the Function

Differentiate the given function \( f(x) = 3x^2 + \frac{2}{x} \). Use the power rule for differentiation: \( \frac{d}{dx}[x^n] = nx^{n-1} \), and the derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \).\[ f'(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}\left(\frac{2}{x}\right) \] \[ f'(x) = 6x - \frac{2}{x^2} \]

Step 3: Evaluate the Derivative at the Given Point

Now, substitute \( x = -2 \) into the derivative \( f'(x) = 6x - \frac{2}{x^2} \).\[ f'(-2) = 6(-2) - \frac{2}{(-2)^2} \] \[ f'(-2) = -12 - \frac{2}{4} \] \[ f'(-2) = -12 - 0.5 = -12.5 \]

Step 4: Conclude Your Findings

You have evaluated the derivative of the function at \( x = -2 \). Hence the value of \( f'(-2) \) is \( -12.5 \).

Key Concepts

Power Rule

The power rule is a fundamental tool in calculus for finding the derivative of polynomial functions. It provides a simple technique to differentiate terms of the form \(x^n\). According to the power rule, if you have a function \( y = x^n \), then its derivative with respect to \( x \) is \( nx^{n-1} \). This means you bring the exponent down as a coefficient and decrease the exponent by one.

Let's illustrate this with an example: Consider the term \(3x^2\). By applying the power rule:

• The coefficient (3) remains the same.
• Apply the power rule to \(x^2\) to get \(2x^{2-1} = 2x\).

Thus, the derivative of \(3x^2\) is \(6x\) (since \(3 \times 2x = 6x\)).

Using the power rule simplifies the differentiation process significantly, especially when dealing with polynomial functions.

Differentiation

Differentiation is the process used in calculus to calculate the derivative of a function. It measures how a function's output value changes as its input value changes. The derivative provides insight into the function's rate of change or slope at any given point.

In the exercise, differentiation was applied to the function \( f(x) = 3x^2 + \frac{2}{x} \). Here's how it works:

• First, differentiate \(3x^2\) using the power rule to get \(6x\).
• Next, rewrite the term \(\frac{2}{x}\) as \(2x^{-1}\) and apply the power rule.
• Using the power rule on \(2x^{-1}\) gives \(-2x^{-2}\), which can be rewritten as \(-\frac{2}{x^2}\).

The overall derivative of the function then becomes \(f'(x) = 6x - \frac{2}{x^2}\). This new function, \(f'(x)\), tells us the slope of the original function \(f(x)\) at any value of \(x\).

Derivative Evaluation

Derivative evaluation involves finding the specific slope or rate of change of a function at a given point. After deriving the function, you substitute a particular x-value into the result to find the derivative at that point.

In this exercise, after differentiating the function to get \(f'(x) = 6x - \frac{2}{x^2}\), we need to evaluate this at \(c = -2\). This is done by substituting \(-2\) for \(x\) in the derivative:

• First, substitute: \(6(-2) - \frac{2}{(-2)^2}\).
• Calculate the terms: \(6(-2)\) becomes \(-12\).
• \(-\frac{2}{(-2)^2}\) simplifies to \(-\frac{2}{4} = -0.5\).
• The final step is combining the results: \(-12 - 0.5 = -12.5\).

Therefore, the derivative evaluated at \(x = -2\) is \(-12.5\), indicating how steeply the function \( f \) is changing at that point.

Functions

Functions are essential mathematical objects in calculus and many areas of mathematics. A function defines a relationship between a set of inputs and a set of possible outputs where each input is related to exactly one output.

In the given exercise, the function \(f(x) = 3x^2 + \frac{2}{x}\) is composed of two terms: a polynomial term \(3x^2\) and a rational term \(\frac{2}{x}\). Each contributes to the behavior of the function over its domain.

• The polynomial part \(3x^2\) dictates the overall shape as \(x\) becomes large or small.
• The rational part \(\frac{2}{x}\) impacts the behavior, especially as \(x\) approaches zero, where it becomes undefined.

Understanding both components helps assess how the function behaves and changes, which is often what differentiation aims to explore. Functions like these are common in calculus, making it essential to become familiar with their properties and behaviors.