Problem 71 - Calculus for Biology and Medicine

Question

In Problems \(71-74\), find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. $$ f(x)=\frac{x^{2}+3}{x^{3}+5}, \text { at } x=-2 $$

Step-by-Step Solution

Answer

The tangent line at \( x = -2 \) is \( y = -8x - \frac{55}{3} \).

Step 1: Find the derivative of the function

To find the tangent line, we first need the derivative of the function, which gives us the slope of the tangent. For the function \( f(x) = \frac{x^2 + 3}{x^3 + 5} \), we use the quotient rule, which is \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \), where \( u = x^2 + 3 \) and \( v = x^3 + 5 \).

Step 2: Apply the quotient rule

Differentiate \( u \) and \( v \): \( u' = 2x \) and \( v' = 3x^2 \). Applying the quotient rule gives:\[f'(x) = \frac{(2x)(x^3 + 5) - (x^2 + 3)(3x^2)}{(x^3 + 5)^2}\] Simplifying, the expression for \( f'(x) \) becomes:\[f'(x) = \frac{2x^4 + 10x - 3x^4 - 9x^2}{(x^3 + 5)^2}\] Further simplifying, we get:\[f'(x) = \frac{-x^4 - 9x^2 + 10x}{(x^3 + 5)^2}\]

Step 3: Evaluate the derivative at the given point

Substitute \( x = -2 \) into the derivative to find the slope of the tangent line at that point. Substitute in the simplified derivative:\[f'(-2) = \frac{-(-2)^4 - 9(-2)^2 + 10(-2)}{((-2)^3 + 5)^2}\]Calculate:\[f'(-2) = \frac{-16 - 36 - 20}{(-8 + 5)^2} = \frac{-72}{9} = -8\]. Thus, the slope \( m \) at \( x = -2 \) is \(-8\).

Step 4: Find the y-coordinate at x = -2

Calculate \( f(-2) \) by substituting \( x = -2 \) into \( f(x) \):\[f(-2) = \frac{(-2)^2 + 3}{(-2)^3 + 5} = \frac{4 + 3}{-8 + 5} = \frac{7}{-3} = -\frac{7}{3}\]So, the point of tangency is \((-2, -\frac{7}{3})\).

Step 5: Use point-slope form to write the equation

The equation of a line with slope \( m \) through a point \((x_1, y_1)\) is \( y - y_1 = m(x - x_1) \). Substituting \( m = -8 \), \( x_1 = -2 \), and \( y_1 = -\frac{7}{3} \), we get:\[y + \frac{7}{3} = -8(x + 2)\] Simplify to slope-intercept form, \( y = mx + b \):\[y = -8x - 16 - \frac{7}{3} = -8x - \frac{48}{3} - \frac{7}{3} = -8x - \frac{55}{3}\]. The equation is \( y = -8x - \frac{55}{3} \).

Key Concepts

Quotient Rule

The Quotient Rule is an essential formula used in calculus to differentiate functions that are ratios of two other functions. When you have a function in the form of \( \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions of \( x \), the quotient rule helps determine the derivative of this quotient.
The rule states that the derivative of \( \frac{u}{v} \) is:

• \( \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^2} \)

In other words, it involves differentiating the numerator (\( u' \)) and the denominator (\( v' \)), multiplying crosswise (\( u'v \) and \( uv' \)), and subtracting these terms before dividing by the square of the original denominator \( v^2 \).
This formula is incredibly useful when dealing with complex fractions in calculus, as manually expanding such expressions would be inefficient.

Derivative Calculation

Calculating derivatives involves finding the rate at which a function changes at any given point, often using various rules of differentiation. In the context of the original exercise, the derivative calculation is crucial for finding the slope of the tangent line.
We start by determining the derivative of the given function using the quotient rule. The function is \( f(x) = \frac{x^2 + 3}{x^3 + 5} \) and we need to find its derivative:

• The numerator, \( u = x^2 + 3 \), has the derivative \( u' = 2x \).
• The denominator, \( v = x^3 + 5 \), has the derivative \( v' = 3x^2 \).
• The quotient rule gives us the derivative \( f'(x) = \frac{(2x)(x^3 + 5) - (x^2 + 3)(3x^2)}{(x^3 + 5)^2} \).

This complicated-looking expression simplifies to \( f'(x) = \frac{-x^4 - 9x^2 + 10x}{(x^3 + 5)^2} \). By evaluating this derivative at the point \( x = -2 \), we find the slope of the tangent line.

Slope-Intercept Form

The slope-intercept form is a way to express the equation of a straight line. It is given as \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
In the original step-by-step solution, once we have the derivative calculated at the desired point, providing a slope of \( -8 \), we use this to write the equation in slope-intercept form.
To do this, we plug in the slope and compute the y-intercept using the coordinates of the point where the tangent touches the curve:

• The slope \( m \) is \(-8 \).
• Substitute into the equation with \( y_1 \) and \( x_1 \) values from the point \((-2, -\frac{7}{3}) \).
• Simplify to get \( y = -8x - \frac{55}{3} \).

This form makes it clear to see how the function height changes with \( x \), making it useful for graphing and understanding the line's behavior.

Point-Slope Form

The point-slope form is another useful method for writing the equation of a line. This is expressed as \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a known point on the line, and \( m \) is the slope.
In many cases, especially when dealing with derivatives and tangent lines, point-slope form is ideal because it directly incorporates a specific point we already know.
With the original problem, once we know the slope \( m = -8 \) and have calculated the exact point of tangency \((-2, -\frac{7}{3}) \), we can form the equation:

• Start with \( y - (-\frac{7}{3}) = -8(x - (-2)) \).
• Simplify to find it equivalent to the equation found in slope-intercept form, \( y = -8x - \frac{55}{3} \).

Unlike the slope-intercept form, point-slope makes it easy to start with specific information like a specific point and the slope, thus paving the way for further simplification into other forms.