Problem 41 - Calculus

Question

The tangent line to \(f(x)=x^{2}-1\) at \(x_{0}=2\) crosses the \(x\) axis at \(x_{1}=\ldots\). The tangent line at \(x_{1}\) crosses the \(x\) axis at \(x_{2}=\) Draw the curve and the two lines, which are the beginning of Newton's method to solve \(f(x)=0\)

Step-by-Step Solution

Answer

The tangent lines cross the x-axis at \( x_1 = \frac{5}{4} \) and \( x_2 = \frac{61}{80} \).

Step 1: Find the Derivative

The function given is \( f(x) = x^2 - 1 \). We need to find the derivative \( f'(x) \) to get the slope of the tangent line. Using the power rule, the derivative is \( f'(x) = 2x \).

Step 2: Evaluate the Derivative at \( x_0 = 2 \)

Calculate the derivative at \( x_0 = 2 \), which is \( f'(2) = 2(2) = 4 \). This gives us the slope of the tangent line at \( x = 2 \).

Step 3: Find the Equation of the Tangent Line

The equation of the tangent line at a point \((x_0, f(x_0))\) is given by \( y - f(x_0) = f'(x_0)(x - x_0) \). Substitute \( x_0 = 2 \) and \( f(x_0) = f(2) = 2^2 - 1 = 3 \) to get \( y - 3 = 4(x - 2) \). Simplifying gives \( y = 4x - 5 \).

Step 4: Find where the Tangent Line Crosses the x-axis

Set \( y = 0 \) in the tangent line equation \( y = 4x - 5 \) and solve for \( x \): \( 0 = 4x - 5 \Rightarrow 4x = 5 \Rightarrow x = \frac{5}{4} \). This point is \( x_1 = \frac{5}{4} \).

Step 5: Compute the Tangent at \( x_1 = \frac{5}{4} \)

Find the slope of the tangent line at \( x_1 = \frac{5}{4} \) using the derivative: \( f'(\frac{5}{4}) = 2 \times \frac{5}{4} = \frac{10}{4} = \frac{5}{2} \). The point on the curve is \( f(\frac{5}{4}) = (\frac{5}{4})^2 - 1 = \frac{25}{16} - 1 = \frac{9}{16} \).

Step 6: Equation of the Tangent Line at \( x_1 \)

Using the point-slope form \( y - f(x_1) = f'(x_1)(x - x_1) \) with \( x_1 = \frac{5}{4} \) and \( f(x_1) = \frac{9}{16} \), the equation becomes \( y - \frac{9}{16} = \frac{5}{2} (x - \frac{5}{4}) \). Simplifying, we get \( y = \frac{5}{2}x - \frac{61}{32} \).

Step 7: Find where the Second Tangent Line Crosses the x-axis

Set \( y = 0 \) in the equation \( y = \frac{5}{2}x - \frac{61}{32} \) and solve for \( x \): \( 0 = \frac{5}{2}x - \frac{61}{32} \). This implies \( \frac{5}{2}x = \frac{61}{32} \), so \( x = \frac{61}{32} \times \frac{2}{5} = \frac{122}{160} = \frac{61}{80} \). This is \( x_2 = \frac{61}{80} \).

Key Concepts

Tangent Lines

A tangent line is a straight line that touches a curve at a single point without crossing through it at that point. Imagine you are standing on a hill at a specific point, and the direction in which you naturally face without turning your head is akin to the tangent line on the curve you're standing on. In the problem given, we are given the curve defined by the function \( f(x) = x^2 - 1 \) and we need to find the tangent line at \( x = 2 \).
The tangent line has the same slope as the curve at the point of tangency. For us to determine the exact equation of this tangent line, we first need to find how steep this curve is at that specific point, which is where the derivative becomes crucial.

Derivative

The derivative of a function at a given point essentially tells us the slope of the tangent line at that point. If you're familiar with skiing down a hill, the derivative at any point on the path will tell you whether you should expect a gentle glide or a steep descent. Calculating the derivative using the power rule allows us to find this slope efficiently.
For the function \( f(x) = x^2 - 1 \), the power rule tells us to bring the exponent down and multiply it by the coefficient. Here, the derivative becomes \( f'(x) = 2x \). This result is essential because it gives us a quick way to calculate the slope of the tangent line at any point, such as at \( x = 2 \). Evaluating the derivative at \( x = 2 \), we find \( f'(2) = 4 \), indicating that the slope of our tangent line at \( x = 2 \) is 4.

Roots of Equations

A root of an equation refers to a solution where the equation equals zero. Just like finding where a ball hits the ground in a game of catch, identifying the roots of an equation involves finding where the function's value is zero on the graph. With Newton's method, which is an iterative process, we start with an initial estimate and use tangent lines to get progressively closer to the actual root.
In the given problem, the roots are the x-values where the tangent lines intersect the x-axis. Initially, we find the root of the first tangent line, where it crosses the x-axis, given by setting \( y = 0 \) in its equation. This process is repeated to find the second intersection of another tangent line with the x-axis, honing in on the solution.

Power Rule

The power rule is an essential tool in calculus for finding derivatives. It provides a shortcut for determining the derivative of a power function, which often acts as the foundation for handling more complex functions. If you have a term like \( x^n \), using the power rule involves bringing down the exponent \( n \) in front of the \( x \) as a multiplier and then reducing the exponent by one.
This rule vastly simplifies working out the derivative of polynomial functions. For example, in \( f(x) = x^2 - 1 \), the power rule quickly gives us \( f'(x) = 2x \), which simplifies our work in finding the slope of tangent lines. This process allows us to understand how functions behave and change, crucially aiding in methods like finding roots through Newton's methodology.