How to Calculate a Linear Approximation

Many students assume linear approximation replaces a curve with its tangent line perfectly. It does not. The tangent line matches the function's value and slope at exactly one point. Move away from that point and the approximation loses accuracy, with error growing roughly as the square of the distance. That is why knowing how to find linear approximation by hand matters: you control the point, you estimate the error, and you decide whether the result is good enough for your problem. The method fails when the curve bends sharply or the point is too far from the target.

Linearization is a five-step process. Every problem follows the same sequence, whether you are approximating a square root, a trig function, or a logarithm.

Step 1: Choose the Function and a Point a

Identify the function f(x) you need to estimate. Pick a point a that is close to your target x-value and where both f(a) and f'(a) are easy to compute. A poor choice of a is the most common cause of large error. The distance |x, a| should be as small as possible.

Step 2: Evaluate f(a)

Plug a into the function. This gives the y-coordinate where the tangent line touches the curve.

Step 3: Find f'(x) and Evaluate at a

Compute the derivative. Then substitute a to get f'(a), the slope of the tangent line at that point. Forgetting to evaluate the derivative at a and leaving it as a function is a frequent mistake.

Step 4: Write the Linearization

The formula is L(x) = f(a) + f'(a)(x, a). This is the tangent line equation.

Step 5: Plug in Your x-value and Simplify

Substitute the x you want to estimate. Perform the arithmetic to get a number. That number is the linear approximation.

After you finish, you can check whether the estimate is an over- or underestimate. If f''(a) is positive (concave up), the tangent line sits below the curve and L(x) is an underestimate. If f''(a) is negative (concave down), L(x) is an overestimate. This concavity test comes from local linearity, the same idea that powers Taylor's inequality for bounding error.

The single most important decision in a linear approximation is the choice of a. A good a makes f(a) and f'(a) trivial to compute. A bad a forces you to compute something as hard as the original problem.

For square roots, use a perfect square near your target. For exponentials, use a = 0 because e⁰ = 1 and the derivative is 1. For sine and cosine, use a = 0 because sin 0 = 0, cos 0 = 1, and the derivatives are simple. For natural logs, use a = 1 because ln 1 = 0 and the derivative is 1.

The rule: the point a must be a number where you know both the function value and the derivative value without a calculator. If you cannot compute f(a) and f'(a) in your head, you chose the wrong a.

Approximate √4.1 using a = 4.

f(x) = √x, so f(4) = 2. f'(x) = 1/(2√x), so f'(4) = 1/(2·2) = 0.25. L(x) = 2 + 0.25(x, 4). For x = 4.1: L(4.1) = 2 + 0.25(0.1) = 2.025. The actual √4.1 is about 2.024845.

Because f''(x) = -1/(4x^(3/2)), which is negative at x = 4, the function is concave down. The linearization is an overestimate.

Approximate (1.02)⁵ using a = 1.

f(x) = x⁵, so f(1) = 1. f'(x) = 5x⁴, so f'(1) = 5. L(x) = 1 + 5(x, 1). For x = 1.02: L(1.02) = 1 + 5(0.02) = 1.10. The actual value is 1.02⁵ ≈ 1.10408. The error is about 0.00408.

f''(x) = 20x³, positive at x = 1, so the function is concave up and the linearization is an underestimate. Moving just 0.02 units away produces a noticeable error because the derivative at a is steep and the second derivative is large.

Approximate sin(0.15) using a = 0. Radian measure is mandatory here. If you use degrees, the small-angle approximation sin θ ≈ θ fails entirely.

f(x) = sin x, so f(0) = 0. f'(x) = cos x, so f'(0) = 1. L(x) = 0 + 1(x, 0) = x. For x = 0.15: L(0.15) = 0.15.

f''(x) =, sin x, which is negative and near zero at a = 0. The linearization is a slight overestimate. For a comparison, sin(0.5) approximated the same way gives L(0.5) = 0.5, but the true value is about 0.479.

Approximate ln(1.08) using a = 1.

f(x) = ln x, so f(1) = 0. f'(x) = 1/x, so f'(1) = 1. L(x) = 0 + 1(x, 1) = x, 1. For x = 1.08: L(1.08) = 0.08. The actual ln(1.08) is about 0.07696. The error is about 0.00304.

f''(x) =, 1/x², negative at x = 1, so the linearization is an overestimate. The natural log curve bends downward, so the tangent line at x = 1 lies above it. Pushing the approximation to x = 1.5 would give L(1.5) = 0.5 against a true value of about 0.405, an error of 0.095, the error bound from Taylor's inequality for n=1 warns you this is coming.

The following question follows the style from the AP Calculus CED Unit 4 (local linearity). You get a function, a point of tangency, and a request to estimate a value and decide if the estimate is an over- or underestimate.

Question

Let f(x) = √(x + 3). Use a linear approximation at a = 1 to estimate f(1.1). Determine whether this approximation overestimates or underestimates the true value.

Solution

f(x) = √(x + 3). f(1) = √4 = 2. f'(x) = 1/(2√(x + 3)). f'(1) = 1/(2·2) = 1/4. L(x) = 2 + (1/4)(x, 1). At x = 1.1: L(1.1) = 2 + (1/4)(0.1) = 2 + 0.025 = 2.025.

f''(x) = -1/(4(x + 3)^(3/2)). At x = 1, f''(1) = -1/(4·8) = -1/32, which is negative. The function is concave down near a = 1, so the tangent line lies above the curve. The linear approximation of 2.025 is an overestimate of f(1.1).

AP-style questions expect you to state the linearization explicitly, compute the estimate, and justify the over/underestimate using the second derivative or concavity. The point of tangency a = 1 works because f(1) and f'(1) are clean integers. A different a would complicate the arithmetic without improving the result.

Function	a Used	x Target	Approximation	True Value	Error
√x	4	4.1	2.025	2.024845	0.000155
x⁵	1	1.02	1.10	1.10408	0.00408
sin x	0	0.15	0.15	0.1482	0.0018
ln x	1	1.08	0.08	0.07696	0.00304