Problem 47 - Brief Applied Calculus

Question

Find the equation for the tangent line to the curve \(y=f(x)\) at the given \(x\) -value. $$ f(x)=(2 x+1)^{4} \text { at } x=-1 $$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = -8x - 7 \).

Step 1: Find the Derivative

To find the tangent line's equation, we first need the derivative of the function, since the slope of the tangent line at a given point is equal to the derivative evaluated at that point. The function given is \( f(x) = (2x+1)^4 \). We use the chain rule to differentiate it:The chain rule states that if \( y = g(h(x)) \), then \( \frac{dy}{dx} = g'(h(x)) \cdot h'(x) \).Here, let \( u = 2x+1 \), thus \( f(x) = u^4 \). We find \( \frac{du}{dx} = 2 \) and \( \frac{d}{dx} u^4 = 4u^3 \). So, the derivative is: \( f'(x) = 4(2x+1)^3 \times 2 = 8(2x+1)^3 \).

Step 2: Evaluate the Derivative at x = -1

Next, we evaluate the derivative at the given point \( x = -1 \):\( f'(x) = 8(2(-1)+1)^3 = 8(-2+1)^3 = 8(-1)^3 = 8(-1) = -8 \).So, the slope of the tangent line at \( x = -1 \) is \(-8\).

Step 3: Find the y-coordinate of the Point on the Curve

Now, we need the point on the curve at \( x = -1 \) to find the equation of the tangent line. We calculate \( f(-1) \):\( f(-1) = (2(-1)+1)^4 = (-2+1)^4 = (-1)^4 = 1 \).Thus, the point on the curve is \((-1, 1)\).

Step 4: Write the Equation of the Tangent Line

With the slope \( m = -8 \) and the point \((-1, 1)\), use the point-slope form of a line \( y-y_1 = m(x-x_1) \):\( y - 1 = -8(x + 1) \).Simplify this to get the equation of the tangent line:\( y - 1 = -8x - 8 \ y = -8x - 7 \).

Key Concepts

Derivative

A derivative serves as a tool in calculus to measure how a function changes as its input changes. Specifically, it provides the slope of the tangent line to the curve at any given point.
It acts as a bridge between algebraic expressions and their corresponding graphical representations.
To find the derivative of a function, you are essentially determining the function's rate of change. In this exercise, the original function is given as \( f(x) = (2x+1)^4 \). To differentiate this function, we introduce a concept called the chain rule, as it involves a composite function.

Chain Rule

The chain rule is a powerful differentiation tool used when dealing with composite functions, or functions within other functions. The rule provides a way to differentiate each part individually and then combine them for the entire function.
In more formal terms, if you have a function \( y = g(h(x)) \), the chain rule states that the derivative \( \frac{dy}{dx} = g'(h(x)) \cdot h'(x) \). In our given exercise, we set \( u = 2x + 1 \) which simplifies the function to \( f(x) = u^4 \). Therefore:

• The inner function \( h(x) = 2x+1 \) has a derivative \( \frac{du}{dx} = 2 \).
• The outer function \( u^4 \) has a derivative \( 4u^3 \).

Combine these to form the overall derivative: \( 4(2x+1)^3 \times 2 = 8(2x+1)^3 \). This derivative gives the slope of the function at any point along the curve.

Point-Slope Form

The point-slope form is an equation of a straight line and is particularly useful when you know a point on the line and the slope of the line. The formula for point-slope form is:\[ y - y_1 = m(x - x_1) \]In this form:

• \( m \) denotes the slope of the line.
• \( (x_1, y_1) \) represents a point through which the line passes.

In this exercise, we found the slope \( m = -8 \) from evaluating the derivative at the point \( x = -1 \). We also calculated the coordinate of the point on the curve \( (-1, 1) \). Substituting these values into the point-slope form, we set up the equation:\( y - 1 = -8(x + 1) \). After simplification, this becomes \( y = -8x - 7 \), the final equation of the tangent line.