Question

Given the function \(f\) and the point \(Q\). find all points \(P\) on the graph of \(f\) such that the line tangent to \(f\) at P passes through \(Q\). Check your work by graphing \(f\) and the tangent lines. $$f(x)=\frac{1}{x} ; Q(-2,4)$$

Short answer

Question: Find the points on the graph of f(x) = 1/x at which the tangent passes through the point Q(-2, 4). Answer: The points P on the graph of f(x) with tangent lines passing through the point Q are given by: $$P_1 (\sqrt{6}, \frac{1}{\sqrt{6}}) \text{ and } P_2 (-\sqrt{6}, -\frac{1}{\sqrt{6}})$$

Step-by-step solution

Find the derivative of f(x)

To find the tangent line at any point P on the graph of the function, we need the derivative of the function, which gives us the slope of the tangent line at any point on the graph. Let's find the derivative of \(f(x)\): Given, $$f(x)=\frac{1}{x}$$ Now differentiate f(x) with respect to x, $$f'(x) = -\frac{1}{x^2}$$

Find the slope of the tangent line passing through Q

To find the slope of the tangent line passing through Q, we need to evaluate the derivative, f'(x), at the x-coordinate of point P. Since we do not know the x-coordinate of P, we cannot find the slope directly. However, we can express it as a function of the x-coordinate of P: Slope of tangent line at P, \(m_{tangent} = f'(x_P) = -\frac{1}{x_P^2}\)

Find the equation of the tangent line

Now, we find the equation of the tangent line using the point-slope form: $$y - y_P = m_{tangent}(x - x_P)$$ Since \(y_P = f(x_P)\), we can write the equation as: $$y - \frac{1}{x_P} = -\frac{1}{x_P^2}(x - x_P)$$

Substitute Q into tangent line equation

We want the tangent line to pass through Q(-2, 4), so substitute the coordinates of Q into the tangent line equation: $$4 - \frac{1}{x_P} = -\frac{1}{x_P^2}(-2 - x_P)$$

Solve for x_P

Simplify the equation and solve for \(x_P\): $$4x_P^2 - x_P^3 - 2x_P = x_P$$ $$x_P^3 - 6x_P = 0$$ $$x_P(x_P^2 - 6)=0$$ There are three possible values for \(x_P\): 0, \(\sqrt{6}\), and \(-\sqrt{6}\). However, since f(x) is defined for all \(x \neq 0\), then \(x_P=0\) is not a valid solution. Thus, there are two valid points P, one with x-coordinate \(\sqrt{6}\) and the other with x-coordinate \(-\sqrt{6}\).

Find the y-coordinates of the points P

Plug the x-coordinates of P into the function f(x) to find the y-coordinates: $$y_P = \frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}$$

Check the results graphically

The points P on the graph of \(f\) with tangent lines passing through Q are given by: $$P_1 (\sqrt{6}, \frac{1}{\sqrt{6}}) \text{ and } P_2 (-\sqrt{6}, -\frac{1}{\sqrt{6}})$$ Plot the graph of the function f(x) and the tangent lines at P1 and P2 to verify that they pass through Q(-2,4).

Key concepts

Derivative of a Function

When we talk about the derivative of a function, we delve into one of the most powerful concepts in calculus. It's the mathematical way to determine the rate at which a function is changing at any given point. Essentially, it's a measure of how a function's output value changes as its input value changes.

In more practical terms, if you imagine a curve on a graph, the derivative at any point on that curve gives you the slope of the line that just touches the curve at that point — this line is known as the tangent. Think of it as a snapshot of the function's behavior in that infinitesimal moment.

To actually calculate the derivative, you have to go through a process of finding the limit of the difference quotient as the change in the input approaches zero. In our exercise, the derivative of the function f(x) = \(\frac{1}{x}\) was found to be \(f'(x) = -\frac{1}{x^2}\). By finding this derivative, we now have a formula that can tell us the slope of the tangent line to the curve at any point x.

Slope of a Tangent Line

The slope of a tangent line to a curve at a particular point is a measure of how steep the line is at that point. If you're standing on a hillside, the slope tells you how inclining or declining the ground is under your feet. In mathematical terms, it's the rise over the run, the change in y over the change in x.

In our example, the function \(f(x) = \frac{1}{x}\) has a slope at any point P defined by its derivative, which we've already calculated to be \(f'(x) = -\frac{1}{x^2}\). To get the slope of the tangent through a specific point Q, you simply plug the x-coordinate of P into this derivative formula. This concept is fundamental because it bridges the gap between a static function and dynamic changes along its graph. With this slope, you're now equipped to write the equation of the tangent line, which will describe its path as it grazes the curve at the point P.

Point-Slope Form Equation

The point-slope form is an equation that represents a line on a graph. Think of it as a recipe for drawing a straight line given certain ingredients: a point through which the line passes and the slope of the line.

Mathematically, the equation goes like this: \(y - y_1 = m(x - x_1)\), where m stands for the slope of the line, and (x_1, y_1) is the point that the line passes through. In our worked example, we've already determined the slope of the tangent line m as \(-\frac{1}{x_P^2}\) and we have our point Q with the coordinates (-2, 4). By plugging m and Q into our point-slope form, we get the equation of the tangent line.

This equation is versatile — you can transform it into other forms like slope-intercept or standard form, depending on your needs. It's a foundational mathematical concept used across science and engineering, not just to describe lines but to model physical phenomena and solve problems.