Problem 81 - Calculus Volume 1

Question

For the following functions, use \(f^{\prime \prime}(x)=\lim _{h \rightarrow 0} \frac{f^{\prime}(x+h)-f^{\prime}(x)}{h}\) to find \(f^{\prime \prime}(x).\) $$f(x)=2-3 x$$

Step-by-Step Solution

Answer

The second derivative \( f^{''}(x) = 0 \).

Step 1: Find the First Derivative

To find the second derivative of the function, we first need to compute its first derivative. The function given is a linear function: \( f(x) = 2 - 3x \). The first derivative, \( f'(x) \), is obtained by differentiating \( f(x) \) with respect to \( x \).Since the derivative of a constant is 0 and the derivative of \( -3x \) is \( -3 \), we have:\[ f'(x) = -3 \]

Step 2: Find the Second Derivative Using the Definition

Now that we have \( f'(x) = -3 \), we use the definition of the second derivative:\[ f^{ ext{''}}(x)= ext{lim}_{h ightarrow 0} \frac{f'(x+h)-f'(x)}{h} \]Substitute \( f'(x) = -3 \) into the limit:\[ f^{ ext{''}}(x)= ext{lim}_{h ightarrow 0} \frac{-3 - (-3)}{h} \]\[ f^{ ext{''}}(x)= ext{lim}_{h ightarrow 0} \frac{0}{h} = 0 \]Therefore, the second derivative is \( f^{ ext{''}}(x) = 0 \).

Key Concepts

First Derivative

Finding the first derivative of a function is a fundamental step in calculus. Derivatives measure how a function changes as its input changes. To find the first derivative of a function, we can use some basic derivative rules:

• The derivative of a constant is zero.
• The derivative of a term like \(ax^n\) follows the power rule: bring down the exponent as a coefficient and reduce the exponent by one. In other words, \(\frac{d}{dx}(ax^n) = nax^{(n-1)}\).

In this example, we are given the function \(f(x) = 2 - 3x\). This is a linear function, straightforward to differentiate.

• The derivative of the constant \(2\) is zero, because constants do not change.
• The derivative of \(-3x\) is \(-3\) because the coefficient of \(x\) is already \(-3\), and the variable is to the power of one.

Thus, the first derivative, which describes the rate of change of \(f(x)\), is constant and equals \(f'(x) = -3\).

Second Derivative

The second derivative is the derivative of the first derivative. It measures how the rate of change itself is changing. In mathematical terms, it's assessing the curvature or concavity of the function's graph.

To find the second derivative, we begin with \( f'(x) = -3 \). Since \(-3\) is a constant, just like in the first derivative where the derivative of a constant is zero, the second derivative here also turns out to be zero.

We can confirm this using the limit definition of the second derivative:

• Given: \( f''(x) = \lim_{h \to 0} \frac{f'(x+h) - f'(x)}{h} \)
• Substitute \( f'(x) = -3 \):
• \( f''(x) = \lim_{h \to 0} \frac{-3 - (-3)}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \)

This tells us the rate of change of the function \(f(x) = 2 - 3x\) is constant, and hence its curvature is zero, indicating a perfectly straight line with no bending.

Limit Definition of Derivative

The limit definition of the derivative is foundational in calculus. It's how we mathematically define the process of differentiation, capturing the idea of instantaneous rate of change.

For the first derivative \( f'(x) \), it is defined as:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]This definition captures how \( f \) changes as we make very small changes to \(x\), showing the slope of the tangent line at any point on \( f(x) \).

For the second derivative, it's about repeating this process on \( f'(x) \), helping us understand not just change but how that change is changing, such as curvature:\[ f''(x) = \lim_{h \to 0} \frac{f'(x+h) - f'(x)}{h} \]
In our exercise, the application of this limit definition shows that since \( f'(x) \) is constant, \( f''(x) \), measuring the change of a constant slope, is zero. This aligns with the idea that a linear function, like \(f(x) = 2 - 3x\), is not bending or curving, hence its second derivative is zero.