Question

Where does the tangent line to \(y=\left(x^{2}+1\right)^{-2}\) at \(\left(1, \frac{1}{4}\right)\) cross the \(x\) -axis?

Short answer

The tangent line crosses the x-axis at \(x=\frac{3}{2}\).

Step-by-step solution

Recall the formula for the tangent line

The tangent line to a curve at a given point is represented by the equation: \[ y - y_1 = m(x - x_1) \]where \((x_1, y_1)\) is the point of tangency and \(m\) is the slope of the tangent line.

Calculate the derivative of the function

To find the slope \(m\) of the tangent line, we first need to calculate the derivative of the function \( y = (x^2 + 1)^{-2} \) with respect to \(x\). Using the chain rule and power rule, the derivative is:\[ y' = -2(x^2 + 1)^{-3} imes 2x = -4x(x^2 + 1)^{-3} \].

Find the slope at the given point

Substitute \(x = 1\) into the derivative to find the slope of the tangent line at the point \((1, \frac{1}{4})\):\[ m = -4(1)((1)^2 + 1)^{-3} = -4 \cdot 2^{-3} = -4 \cdot \frac{1}{8} = -\frac{1}{2} \].

Write the equation of the tangent line

Using the point-slope form and the calculated slope, plug \((x_1, y_1) = (1, \frac{1}{4})\) and \(m = -\frac{1}{2}\) into the tangent line equation:\[ y - \frac{1}{4} = -\frac{1}{2}(x - 1) \]Simplify to find the equation of the tangent line:\[ y = -\frac{1}{2}x + \frac{1}{2} + \frac{1}{4} = -\frac{1}{2}x + \frac{3}{4} \].

Find where the tangent line crosses the x-axis

To find where the tangent line crosses the \(x\)-axis, set \(y = 0\) in the tangent line equation:\[ 0 = -\frac{1}{2}x + \frac{3}{4} \]Solve for \(x\):\[ \frac{1}{2}x = \frac{3}{4} \]\[ x = \frac{3}{4} \times 2 \]\[ x = \frac{3}{2} \].

Key concepts

Calculus

Calculus is a fundamental branch of mathematics that studies change. It deals with rates of change and accumulation of quantities. One of its core concepts is differentiation, which focuses on finding the rate at which a function changes at any given point.
The tangent line problem involves calculus because we use differentiation to find the slope of a line that just touches a curve at one point and doesn't cross it. This process allows us to determine how steep the curve is at that point.
In the exercise given, the function is a curved line, making calculus essential to finding exactly where the line touches.

Derivative Calculation

Calculating the derivative is a crucial part of solving tangent line problems. Derivatives provide the slope of a curve at any given point on that curve.
For the function given in the exercise, we need to apply rules of differentiation to find the derivative. Using the chain rule and the power rule, we find that the derivative of the function \( y = (x^2 + 1)^{-2} \) is:

• Derivative: \( y' = -4x(x^2 + 1)^{-3} \)

This derivative tells us how the function's value changes concerning a small change in \(x\).
By substituting \(x = 1\), we find that the slope at that point is \(-\frac{1}{2}\).

Slope of Tangent Line

The slope of a tangent line tells us how steep the line is at a certain point. In calculus, finding this slope is paramount to solving many kinds of problems involving curves.
In our exercise, once the derivative is calculated as \(y' = -4x(x^2 + 1)^{-3}\), the next step is to evaluate this at the given point \((1, \frac{1}{4})\).

• The calculated slope at \(x = 1\): \(-\frac{1}{2}\)

This slope symbolizes the steepness of the line at the point where the curve touches the tangent line. A slope of \(-\frac{1}{2}\) means for every unit we move right on the x-axis, the line moves down half a unit.

Point-Slope Form

The point-slope form is a way of writing the equation of a straight line when you know a point on the line and its slope.
The formula for point-slope form is: \(y - y_1 = m(x - x_1)\). Here, \((x_1, y_1)\) is the known point on the line, and \(m\) is the slope.
Using the exercise's data with point \((1, \frac{1}{4})\) and slope \(-\frac{1}{2}\), we construct the tangent line equation:

• Equation: \(y - \frac{1}{4} = -\frac{1}{2}(x - 1)\)
• This simplifies to: \(y = -\frac{1}{2}x + \frac{3}{4}\)

Once simplified, this equation gives a precise linear representation of the curve at the point \((1, \frac{1}{4})\), and lets us see where it crosses the x-axis. Here, setting \(y = 0\) allows us to solve for \(x = \frac{3}{2}\), the point where the line crosses the axis.