Problem 50 - College Algebra and Calculus: An Applied Approach

Question

Find an equation of the tangent line to the graph of the function at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. $$ y=x^{3}+x \quad(-1,-2) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the given function at the point (-1, -2) is \(y = 4x + 6\).

Step 1: Compute the Derivative

Differentiate the given function \(y = x^{3} + x\). Using the power rule, the derivative \(y' = 3x^{2} + 1\). This derivative function gives the slope of the tangent line at any point on the graph of the function.

Step 2: Substitute the Given Point

Substitute the given point (-1, -2) into the derivative equation to determine the slope of the tangent at that point. Thus, the slope \(m = 3(-1)^{2} + 1 = 3 + 1 = 4\).

Step 3: Use the Slope-Point Form

Now use the slope-point form of the line equation, which is \(y - y_{1} = m(x - x_{1})\), where \((x_{1}, y_{1})\) is the point and \(m\) is the slope. Substituting our given point and computed slope into this equation's we get \(y - (-2) = 4(x - (-1))\), which simplifies to \(y = 4x + 6\).

Step 4: Confirm the Results

Verify your results using a graphing utility. The tangent line equation should match the line drawn on the graph at the point (-1, -2). Also confirm that the derivative at that point matches the slope you calculated, which is 4.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus. They measure how a function changes at any given point. When you calculate the derivative of a function, you’re finding the function's slope at any point along its curve. This is incredibly useful for determining the steepness of a tangent line to a curve. A tangent line touches the curve at exactly one point, and its slope at that point is given by the derivative.

In our example with the function \(y = x^3 + x\), the derivative is computed to be \(y' = 3x^2 + 1\). This derivative tells us the slope of the tangent line for any \(x\) value. When we substitute \(x = -1\) into the derivative, we find the slope at that specific point, which turns out to be 4.

Slope-Point Form

The slope-point form is a particular formula for finding the equation of a line when you know one point on the line and the slope. This formula is expressed as:

• \(y - y_1 = m(x - x_1)\)

Where \((x_1, y_1)\) is the point, and \(m\) is the slope. This form is extremely handy for writing the equation of a tangent line because you can straight away insert the calculated slope and given point into the formula to get the line equation.

In our example, we determined the slope \(m = 4\) at the point \((-1, -2)\). By placing these values into the slope-point form, we derived the tangent line equation: \(y - (-2) = 4(x - (-1))\). Simplifying this gives us \(y = 4x + 6\).

Power Rule

The power rule is a fundamental technique used in calculus to find derivatives quickly and easily. It states that if you have a function of the form \(x^n\), where \(n\) is a real number, the derivative is given by:

• \( \frac{d}{dx}(x^n) = nx^{n-1} \)

This simple rule allows us to compute derivatives without having to return to the definition of a derivative every time.

In the original exercise, the function \(y = x^3 + x\) was differentiated using the power rule. For \(x^3\), applying the power rule gives us \(3x^2\). The derivative of \(x\) can also be found using the rule, considering it as \(x^1\), which simply results in 1.

Thus, the derivative of the function \(y = x^3 + x\) is \(y' = 3x^2 + 1\), highlighting how the power rule facilitates the process of differentiation.