Problem 10 - A Graphical Approach to Precalculus with Limits

Question

Find the slope of the tangent line to each curve when \(x\) has the given value. Do not use a calculator. $$f(x)=\frac{6}{x} ; x=-1$$

Step-by-Step Solution

Answer

The slope of the tangent line is -6.

Step 1: Find the Derivative

To find the slope of the tangent line, we first need the derivative of the function. Our function is \( f(x) = \frac{6}{x} \). This is the same as \( f(x) = 6x^{-1} \). To differentiate, use the power rule which states that \( \frac{d}{dx}x^n = nx^{n-1} \). Applying the power rule gives us the derivative: \( f'(x) = -6x^{-2} \).

Step 2: Evaluate the Derivative at Given Point

Now that we have the derivative, we need to evaluate it at the given point, \( x = -1 \). Substitute \( x = -1 \) into the derivative: \( f'(-1) = -6(-1)^{-2} \).

Step 3: Simplify the Expression

We need to simplify the expression \( -6(-1)^{-2} \). Since \( (-1)^{-2} = \frac{1}{(-1)^2} = \frac{1}{1} = 1 \), the expression simplifies to \( -6 \times 1 \). Thus, \( f'(-1) = -6 \).

Key Concepts

Power Rule

The Power Rule is a fundamental concept in calculus that is all about taking derivatives of functions with ease. When you have a function in the form of \( x^n \), where \( n \) is any real number, the Power Rule comes into play. It states that the derivative of \( x^n \) is \( n \times x^{n-1} \). This makes it very handy for differentiation.

In our exercise, the function given was \( f(x) = \frac{6}{x} \). This can be rewritten using exponents as \( f(x) = 6x^{-1} \). Here, the exponent \( n \) is -1. To find the derivative, apply the Power Rule. Multiply the constant (6) by the exponent (-1), giving \( -6 \). Then, reduce the exponent by one: \( -1 - 1 = -2 \). So, the derivative \( f'(x) = -6x^{-2} \).

• Start by identifying the function format
• Apply the rule: derivative of \( x^n \) is \( n \times x^{n-1} \)
• Simplify the expression

Slope of Tangent Line

The slope of a tangent line to a curve at any given point tells us how steep the curve is at that exact point. In calculus, the derivative of a function gives you a new function that describes the slope of the tangent line at any \( x \) value you might choose.

When working with our example, after we found the derivative \( f'(x) = -6x^{-2} \), this expression represents the slope of the tangent line to \( f(x) \). Tangent lines use this slope to tell us whether a curve is heading upwards or downwards and by how much. In simpler terms, it snaps a straight, touching line to that point on the curve, representing instantaneous change.

• The derivative function gives you the slope at any point
• For a given \( x \), substitute it into the derivative to find the slope
• Tangent lines visualize a curve's rate of change precisely

Evaluation at a Point

Evaluating a derivative at a specific point is crucial for finding the actual slope of the tangent line right there. It's straightforward but vital in understanding how the tangent line relates to the curve at that point.

For the problem at hand, once we've determined the derivative \( f'(x) = -6x^{-2} \), the task is to plug in the given \( x \) value, which was \( x = -1 \). Substituting \( x = -1 \) into the derivative gives \( f'(-1) = -6(-1)^{-2} \). Simplifying this, we find that \((-1)^{-2} = 1\), so \( f'(-1) = -6 \times 1 = -6 \). Thus, the slope of the tangent line at \( x = -1 \) is \(-6\).

• Find the derivative expression first
• Substitute the specific \( x \) value
• Simplify the expression for the exact slope