A Tangent Line Problem

The problem: Find k such that the line is tangent to the graph of the function f(x) = k/x.

Thinking about the solution: If the line is tangent to the graph at some point (x,y), then there are two conditions that must be met:

1. that point must satisfy both equations, and
2. the slope of f(x) at that point must equal the slope of the line.

Since it’s not immediately obvious what the solution is, let’s write both conditions in algebra and see what may come up.

1. If a point (x,y) satisfies both equations, then and y = k/x. Combining them (eliminating y) gives .

2. The slope of the line is –¾. The slope of the curve at any x is f′(x) = –k/x². Since those must be equal, you have –¾ = –k/x².

Now we have two equations in k and x. That should be enough to find k, which is what the problem asks for. We really want k and don’t much care about x, but it looks a little easier to solve by eliminating k.

From –¾ = –k/x² get k = 3x²/4. Substitute in the other equation and we have . Simplify the right-hand side to get . This is easily solved to get x = 2. Now substitute x = 2 back in k = 3x²/4 to get k = 3.

Check: Our answer is that k = 3, but we can prove that that’s correct — and we should! The check is to make sure that our solution satisfies both of the numbered criteria for a tangent line at a point.

1. We know that x=2 at the point of tangency. If k = 3 then f(2) = 3/2, and the point (2, 3/2) is on the curve. Substituting (2, 3/2) in shows that the line also contains that point.
2. If k = 3, then f′(2) = –¾, which matches the slope of the line.

Conclusion: If k = 3, the graph of f(x) = k/x is tangent to the line at the point (2, 3/2). You might want to graph y = 3/x and to help you visualize the point of tangency.