Build a tangent-line approximation

Zoom in on a curve

The curve is \(y = x^3\). The orange point is \((2, 8)\), and the orange line that touches the curve there is the tangent line.

The tangent line has slope 12, because the derivative \(3x^2\) equals 12 at \(x = 2\).

Move the slider to the left to narrow the window around the point. The red segment is the gap between the curve and the line at \(x = 2 + w\), a distance \(w\) to the right of the orange point.

The gap is 7 when w is 1, about 1.6 when w is 0.5, and about 0.06 when w is 0.1. It shrinks much faster than w, so in a narrow window the curve looks like a straight line.

Near the point, the tangent line can replace the curve, and a line is easier to work with because its slope is constant.

Any differentiable function

The curve is any function \(f\) that has a derivative at \(a\). The orange point is \((a, f(a))\), and the tangent line through it has slope \(f'(a)\).

Move the slider to change the input \(x\). The orange dot shows the height of the line, \(L(x)\), and the chalk dot shows the height of the curve, \(f(x)\).

The red gap between the two dots is small when x is near a and grows as x moves away from a.

The line needs only two numbers: the value \(f(a)\) and the slope \(f'(a)\).

Notation

1. The tangent line passes through \((a, f(a))\) and has slope \(f'(a)\), so in point-slope form its equation is the one below.

   \[y - \cy{f(a)} = \cy{f'(a)}\,(x - a)\]

2. Solving for \(y\) gives a function of \(x\). Call it \(L\), the linearization of \(f\) at \(a\).

   \[L(x) = \cy{f(a)} + \cy{f'(a)}\,(x - a)\]

   The line starts at the height \(f(a)\) and rises by \(f'(a)\) for each unit that \(x\) moves from \(a\). The input \(a\) is called the center.

3. Near \(a\), the line and the curve are close, so \(L(x)\) approximates \(f(x)\).

   \[f(x) \approx L(x) \quad \text{for } x \text{ near } a\]

   The approximation needs \(f'(a)\) to exist, which means that \(f\) is differentiable at \(a\).

When it fails

Put a corner at \(a\), so that the curve has one slope on the left and another on the right. Turn the line to each slope in turn.

Each line follows the curve on one side of the corner but not on the other. No line is tangent on both sides, so \(f'(a)\) does not exist and there is no linearization.

Even when \(f'(a)\) exists, the line is close to the curve only near \(a\). Move the marker far from \(a\), and the red gap grows.

Example 1: The tangent function

Find the linearization of \(f(x) = \tan x\) at \(a = \pi/4\).

1. Find the value at the center.

   \[\cy{f(\pi/4)} = 1\]

2. Find the slope. The derivative of \(\tan x\) is \(\sec^2 x\), and \(\sec^2(\pi/4) = 1/\cos^2(\pi/4) = 2\).

   \[\cy{f'(\pi/4)} = 2\]

3. Substitute both numbers into the linearization.

   \[L(x) = 1 + 2\left(x - \frac{\pi}{4}\right)\]

Example 2: One over x

Find the linearization of \(f(x) = 1/x\) at \(a = 2\).

1. Find the value and the slope at the center. The derivative of \(1/x\) is \(-1/x^2\).

   \[f(2) = \tfrac12, \qquad f'(2) = -\tfrac14\]

2. A tempting wrong move: Write the line with \(x\) in place of \(x - a\).

   \[L(x) = \tfrac12 - \tfrac14 x\]

   At \(x = 2\) this line has height \(\tfrac12 - \tfrac12 = 0\), but the curve is at \(\tfrac12\), so the red line misses the orange point. A tangent line must touch the curve at \(a\).

3. Use \(x - 2\), so that the slope term is 0 at \(x = 2\) and the line passes through \((2, \tfrac12)\).

   \[L(x) = \tfrac12 - \tfrac14(x - 2) = 1 - \tfrac{x}{4}\]

Practice

Find the linearization of \(g(x) = \ln x\) at \(a = e\).

The line must touch the curve at \(x = e\). Find the height and the slope of the curve there.

Questions

Questions 1 and 2 use the curve of \(h(x) = x^2 - 4x + 5\). The left edge of the figure is the y-axis. The point of tangency is \((3, 2)\), and the tangent line there has slope 2.

1. Which formula gives the linearization of \(h\) at \(x = 3\)?

   • \(L(x) = 2 + 2(x - 3)\)
   • \(L(x) = 2 + 3(x - 2)\)
   • \(L(x) = 2 + 2x\)
   Answer

   \(L(x) = 2 + 2(x - 3)\)

2. Where does the tangent line cross the y-axis?

   • At \(y = -4\)
   • At \(y = 5\)
   • At \(y = 2\)
   Answer

   At \(y = -4\)

3. Find the linearization of \(f(x) = \sqrt{3x + 1}\) at \(a = 1\).

   • \(L(x) = 2 + \tfrac34(x - 1)\)
   • \(L(x) = 2 + \tfrac14(x - 1)\)
   • \(L(x) = 2 + \tfrac32(x - 1)\)
   • \(L(x) = 2 + \tfrac34 x\)
   Answer

   \(L(x) = 2 + \tfrac34(x - 1)\)

Summary

You build the linearization \(L\) from the value \(f(a)\) and the slope \(f'(a)\) at the center \(a\).

It works because a differentiable curve looks like its tangent line near a, so L(x) is close to f(x) there.

You choose a center where \(f(a)\) and \(f'(a)\) are easy to find and that is near the input you need.

This is one of four skills in topic 4.6 of Which & Why: AP Calculus.