Linear Approximation of sqrt(x+1) at x=3: Example
The problem: find the linear approximation of f(x) = √(x + 1) at x = 3, and use it to estimate √4.1.
Students find this topic slippery because it asks you to deliberately replace a function with a wrong one: a straight line pretending to be a curve. The discomfort fades once you see the point. Near the base point, a smooth curve and its tangent line are nearly indistinguishable, so the line gives excellent estimates with only arithmetic. The whole method is one formula, L(x) = f(a) + f′(a)(x − a), applied carefully. This example works it end to end and then tests the estimate against the true value.
Worked solution
We want the linearization of f(x) = √(x + 1) at a = 3.
Step 1. Evaluate the function at the base point. f(3) = √(3 + 1) = √4 = 2. The tangent line must pass through the point (3, 2). ✓
Step 2. Differentiate. Write f(x) = (x + 1)^(1/2) and use the power rule with the chain rule:
f′(x) = (1/2)(x + 1)^(−1/2) · 1 = 1 / (2√(x + 1))
Step 3. Evaluate the derivative at the base point. f′(3) = 1 / (2√4) = 1 / (2 · 2) = 1/4. The tangent line has slope 1/4. ✓
Step 4. Assemble the linearization. Plug into L(x) = f(a) + f′(a)(x − a):
L(x) = 2 + (1/4)(x − 3)
Step 5. Simplify. Distribute and collect: L(x) = 2 + x/4 − 3/4 = x/4 + 5/4, or in decimals:
L(x) = 0.25x + 1.25
Step 6. Use it. To estimate √4.1, note that 4.1 = 3.1 + 1, so we need f(3.1):
L(3.1) = 2 + (1/4)(3.1 − 3) = 2 + (1/4)(0.1) = 2 + 0.025 = 2.025
The true value is √4.1 ≈ 2.0248457, so our estimate 2.025 is off by only about 0.0002. That is the power of the method: one line of arithmetic lands within two ten-thousandths of the truth.
Why this method works
Zoom in close enough on any smooth curve and it looks like a straight line; that is what “differentiable” means visually. The tangent line is the unique straight line that matches both the function’s value and its slope at the base point, so near that point the line and the curve agree to first order. The farther you wander from the base point, the more the curve’s bending matters and the worse the estimate gets, which is why linear approximation is a local tool.
Common mistakes
- Using the wrong base point in the formula. L(x) = f(a) + f′(a)(x − a) needs f and f′ evaluated at a = 3, but (x − a) keeps the variable x. Plugging x = 3.1 into f′ instead of keeping f′(3) = 1/4 destroys the estimate.
- Forgetting the chain rule when differentiating. Writing f′(x) = 1/(2√x) drops the +1 inside the root. The derivative of √(x+1) is 1/(2√(x+1)); the inside function x + 1 differentiates to 1, which is invisible but still required.
- Expecting the approximation to work far away. L(10) = 0.25(10) + 1.25 = 3.75 versus √11 ≈ 3.317. Seven units from the base point, the curve has bent away from its tangent. Linear approximation is for nearby x only.
Check (never skip)
Two checks. First, the line must touch the curve at the base point: L(3) = 0.25(3) + 1.25 = 0.75 + 1.25 = 2 = f(3). ✓ Second, the slopes must match: our line rises 0.25 per unit of x, and f′(3) = 1/4 = 0.25. ✓ A line through (3, 2) with slope 1/4 is exactly the tangent line, so L(x) = 0.25x + 1.25 is correct. As a final reality check, L(2.9) = 2 − 0.025 = 1.975 against the true √3.9 ≈ 1.9748: still excellent on the other side of 3. ✓
Frequently Asked Questions
Q: Why does the tangent line approximate the curve?
A: Differentiability means the curve looks straight under enough magnification. The tangent line matches the curve’s position and direction at the base point, so within a small neighborhood the two are practically the same.
Q: When is linear approximation a bad idea?
A: Far from the base point, where the curve’s bending accumulates; where the second derivative is large, meaning the curve bends sharply; or where the function is not differentiable at all, like a corner or cusp.
Q: Can I use a different base point, like a = 0?
A: Yes, and you should pick the base point closest to your target. For √4.1, a = 3 beats a = 0 because 3.1 is nearer to 3. The formula works at any a where f is differentiable; nearer is better.
Try it with steps on our linear approximation calculator.