Problem 116 - Calculus and its applications

Question

Graph fand \(f^{\prime}\) and then determine \(f^{\prime}(1) .\) $$ f(x)=\frac{4 x}{x^{2}+1} $$

Step-by-Step Solution

Answer

The value of \(f'(1)\) is 0.

Step 1: Differentiate the Function

Before we can graph both functions, we need to find the derivative of the given function \(f(x) = \frac{4x}{x^2+1}\). We will use the quotient rule for derivatives, which is \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}\), where \(u = 4x\) and \(v = x^2 + 1\). Their derivatives are \(u' = 4\) and \(v' = 2x\). So, \(f'(x) = \frac{(x^2 + 1)(4) - (4x)(2x)}{(x^2 + 1)^2}\).

Step 2: Simplify the Derivative

Now, simplify the expression obtained from differentiating: \(f'(x) = \frac{4x^2 + 4 - 8x^2}{(x^2 + 1)^2}\), which simplifies to \(f'(x) = \frac{-4x^2 + 4}{(x^2 + 1)^2}\). Further simplification gives \(f'(x) = \frac{4(1-x^2)}{(x^2 + 1)^2}\).

Step 3: Evaluate the Derivative at a Specific Point

Substitute \(x = 1\) into the simplified derivative to find \(f'(1)\). Substitute into the expression: \(f'(x) = \frac{4(1-1^2)}{(1^2 + 1)^2} = \frac{4(1-1)}{2^2} = \frac{0}{4} = 0\). Thus, \(f'(1) = 0\).

Step 4: Graph Function \(f(x)\) and \(f'(x)\)

Graph \(f(x) = \frac{4x}{x^2+1}\) and its derivative \(f'(x) = \frac{4(1-x^2)}{(x^2 + 1)^2}\). The graph of \(f(x)\) is a rational function, showing dynamics between \(x\) and \(f(x)\) values, while the graph of \(f'(x)\) shows where the function increases, decreases, and its slope. At \(x=1\), the derivative is zero, indicating a horizontal tangent.

Key Concepts

Quotient Rule

The Quotient Rule is a essential technique for finding the derivative of a function expressed as a quotient of two differentiable functions. In our exercise, we are calculating the derivative of the function \( f(x) = \frac{4x}{x^2+1} \). Here,

• \(u\) is the numerator and equals 4x.
• \(v\) is the denominator and equals \(x^2 + 1\).

These assignments simplify the differentiation process.
The derivative of \(u\) is \(u' = 4\), and the derivative of \(v\) is \(v' = 2x\).
Using the quotient rule formula\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2},\]we substitute to obtain:\[f'(x) = \frac{(x^2 + 1)(4) - (4x)(2x)}{(x^2 + 1)^2}.\]Simplifying this gives us the derivative in a usable form. This complete step paves the way for further analysis and provides insight into the function's behavior.

Function Graphing

Graphing the original function \( f(x) = \frac{4x}{x^2+1} \) and its derivative \( f'(x) \) provides important visual insights.

• The graph of \( f(x) \) is a rational function and reflects the interplay between the numerator \(4x\) and the denominator \(x^2+1\).
• The function displays certain behaviors such as horizontal asymptotes as \(x\) approaches infinity.

Graphing the derivative \( f'(x) = \frac{4(1-x^2)}{(x^2 + 1)^2} \) helps determine where the function's slope is increasing, decreasing, or staying constant.
By examining this graph, you can easily identify critical points in the original function and understand its overall behavior.
With these graphs, students can clearly see where the function is heading and predict behavior at different values of \(x\).

Slope of the Tangent Line

The slope of the tangent line to a function at a specific point is given by the derivative at that point. In our exercise, we are particularly interested in finding the slope at \(x = 1\).
After already simplifying the derivative to \( f'(x) = \frac{4(1-x^2)}{(x^2 + 1)^2} \), we substitute \(x = 1\) to evaluate the slope at this point:\[f'(1) = \frac{4(1-1^2)}{(1^2 + 1)^2} = \frac{4(1-1)}{2^2} = \frac{0}{4} = 0.\]When \( f'(1) = 0 \), it shows that the slope of the tangent to the function at \(x=1\) is horizontal.
This means that at this specific point, the function has a local extremum, which in simpler terms typically implies a peak or trough.
Understanding the slope assists in sketching the function's curve and anticipating its changes in direction and magnitude of increase or decrease.