Problem 71 - Calculus for Biology and Medicine

Question

Find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. \(f(x)=\frac{x^{2}+3}{x^{3}+5}\), at \(x=0\)

Step-by-Step Solution

Answer

The tangent line at \( x = 0 \) is \( y = \frac{3}{5} \).

Step 1: Find the derivative of the function

To find the tangent line, we need the derivative of the function, which is the slope of the tangent. The function is given by \( f(x) = \frac{x^2 + 3}{x^3 + 5} \). To differentiate, we'll use the quotient rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \), where \( u = x^2 + 3 \) and \( v = x^3 + 5 \). Therefore, \( u' = 2x \) and \( v' = 3x^2 \). The derivative is \( f'(x) = \frac{(x^3 + 5)(2x) - (x^2 + 3)(3x^2)}{(x^3 + 5)^2} \).

Step 2: Simplify the derivative expression

Now, let's simplify \( f'(x) = \frac{2x(x^3 + 5) - 3x^2(x^2 + 3)}{(x^3 + 5)^2} \). This gives \( f'(x) = \frac{2x^4 + 10x - 3x^4 - 9x^2}{(x^3 + 5)^2} \). Combining like terms, we have \( f'(x) = \frac{-x^4 - 9x^2 + 10x}{(x^3 + 5)^2} \).

Step 3: Evaluate the derivative at the given point

We need to find the slope at \( x = 0 \). Substitute \( x = 0 \) into the derivative: \( f'(0) = \frac{-0^4 - 9(0)^2 + 10(0)}{(0^3 + 5)^2} = \frac{0}{25} = 0 \). Thus, the slope is zero.

Step 4: Find the y-coordinate of the given point

The equation of the tangent line requires a point on the function. Substitute \( x = 0 \) into the original function \( f(x) \): \( f(0) = \frac{0^2 + 3}{0^3 + 5} = \frac{3}{5} \). So, the point is \( (0, \frac{3}{5}) \).

Step 5: Write the equation of the tangent line

With the slope \( m = 0 \) and the point \( (0, \frac{3}{5}) \), the equation of the tangent line in point-slope form is \( y - \frac{3}{5} = 0(x - 0) \). Simplifying, the equation becomes \( y = \frac{3}{5} \).

Key Concepts

Understanding the Derivative

The derivative of a function measures how the function's output changes as its input changes. In other words, it tells us the rate of change or the slope of the function at any given point. For our function, knowing the derivative is crucial to finding the equation of a tangent line.

A derivative is found using some specific rules that depend on how the function is structured. One common method is the quotient rule, which we use when the function is a division of two different expressions. In our case, the function is \( f(x) = \frac{x^2 + 3}{x^3 + 5} \). To find the slope of the tangent line to this function at a particular point, we need to compute its derivative.

Applying the Quotient Rule

When working with derivatives of fractions, the quotient rule is your friend! It helps when you have a function represented as the quotient of two other functions. The rule states:

• If \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of their quotient \( \frac{u}{v} \) is \( \frac{v \cdot u' - u \cdot v'}{v^2} \).

To use this rule:

• Identify \( u \) and \( v \). Here, \( u = x^2 + 3 \) and \( v = x^3 + 5 \).
• Find the derivatives of these functions: \( u' = 2x \) and \( v' = 3x^2 \).

When applied, it helps us find the slope at any point by providing the derivative of the combined function. Simplifying the expression at the point of interest gives you the exact slope of the tangent line.

Formatting in Slope-Intercept Form

Once you have the derivative and know the slope at the specific point, the next step is to find the tangent line. Typically, we look for the tangent line in the slope-intercept form: \( y = mx + b \).

• The \( m \) stands for the slope, which is the derivative evaluated at the point.
• The \( b \) is the y-intercept, which we find by plugging our point of interest into the equation of the line.

In our exercise, after calculating the derivative and determining it at \( x = 0 \), we find \( m = 0 \). This simplifies the equation to a horizontal line given by \( y = \frac{3}{5} \) since the slope is zero and the line crosses the y-axis at \( \frac{3}{5} \).

Embracing Calculus

Calculus is a powerful mathematical tool that allows us to analyze functions, understand change, and predict future behaviors. It is divided mainly into two branches: differential calculus (concerned with curves and slopes) and integral calculus (concerned with areas and accumulations).

In this exercise, we're focusing on differential calculus to find a derivative and determine the tangent line to a curve at a particular point. The broader understanding of calculus helps solve real-world problems where changes and trends need to be understood over continuous intervals. For students tackling functions and derivatives, the core takeaway is how these concepts allow for precise definitions of instantaneous rates of change and positions, leading to accurate predictions and analysis.