Question

Find the equations of the tangent and normal lines to the graph of the function at the given point. \(f(x)=x^{3}-x\) at \(x=1\).

Short answer

Tangent: \(y = 2x - 2\); Normal: \(y = -\frac{1}{2}x + \frac{1}{2}\).

Step-by-step solution

Find the function derivative

First, find the derivative of the function to get the slope of the tangent line. The function is \(f(x) = x^3 - x\). Use the power rule \(\frac{d}{dx}x^n = nx^{n-1}\). The derivative is \(f'(x) = 3x^2 - 1\).

Calculate the slope of the tangent line at the point

Substitute \(x = 1\) into the derivative to find the slope of the tangent line at that point. \(f'(1) = 3(1)^2 - 1 = 3 - 1 = 2\). So, the slope of the tangent line is 2.

Find the y-coordinate of the point on the function

Calculate \(f(1)\) to determine the y-coordinate of the point on the graph. \(f(1) = (1)^3 - 1 = 0\). So, the point is \((1, 0)\).

Write the equation of the tangent line

Use the point-slope form to write the equation of the tangent line: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point. The equation becomes \(y - 0 = 2(x - 1)\). Simplify to get the tangent line equation: \(y = 2x - 2\).

Calculate the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is 2, the slope of the normal line is \(-\frac{1}{2}\).

Write the equation of the normal line

Use the point-slope form to write the equation of the normal line: \(y - y_1 = m(x - x_1)\) where \(m\) is the slope of the normal line, and \((x_1, y_1)\) is the point. The equation becomes \(y - 0 = -\frac{1}{2}(x - 1)\). Simplify to get the normal line equation: \(y = -\frac{1}{2}x + \frac{1}{2}\).

Key concepts

Derivative

A derivative in calculus represents the rate at which a function is changing at any point. Essentially, it tells us the slope of the tangent line to the function's curve at any given point. In the exercise, we worked with the function \( f(x) = x^3 - x \). To find its derivative, we applied a simple rule called the Power Rule, which helps us determine how each term in the polynomial contributes to the slope. The derivative of the function, \( f'(x) = 3x^2 - 1 \), provides us with crucial information about the behavior of the function: it defines how the slope changes across different values of \( x \). By evaluating this derivative at a specific point, like \( x = 1 \), we obtained the slope of the tangent line at that point, which turned out to be 2. Understanding derivatives is key for exploring concepts like tangent and normal lines, as they allow us to describe how fast functions are changing at particular instances.

Point-Slope Form

The point-slope form of a line's equation is an efficient way to construct equations for lines when we know a point on the line and its slope. The formula is expressed as \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a given point, and \( m \) is the slope of the line.

In our task, once we had the point \( (1, 0) \) and the slope found from the derivative, this form allowed us to quickly write the equation of the tangent line at that point. Substituting \( m = 2 \), the formula became \( y - 0 = 2(x - 1) \), which simplifies to \( y = 2x - 2 \).

Later, to find the normal line, which is perpendicular to the tangent line, we calculated the negative reciprocal of the tangent slope, \( -\frac{1}{2} \), and used the same point to find its equation: \( y - 0 = -\frac{1}{2}(x - 1) \), simplifying to \( y = -\frac{1}{2}x + \frac{1}{2} \). Using point-slope form is versatile and straightforward, enabling the derivation of line equations easily from known data.

Power Rule

The Power Rule is a quick and convenient shortcut for finding derivatives of polynomial functions. It states that if you have a term in the form \( x^n \), its derivative is \( nx^{n-1} \). This rule simplifies the process by allowing us to differentiate terms individually within a polynomial.

In the exercise, the function \( f(x) = x^3 - x \) consists of two terms: \( x^3 \) and \( -x \). For \( x^3 \), applying the Power Rule gives \( 3x^2 \), and for \( -x \), we recall that \( x \) is the same as \( x^1 \), leading to a derivative of \( 1 \cdot x^0 = 1 \). Therefore, the derivative \( f'(x) = 3x^2 - 1 \) was derived directly from applying this fundamental rule.

The Power Rule greatly simplifies derivative calculations, turning what could be a laborious algebraic process into manageable and quick computations. It's a fundamental tool in calculus that students regularly use to analyze and interpret the behavior of polynomial functions by measuring how they change over their domains.