Difficulty: Intermediate | Prerequisites: Derivatives (power rule, chain rule, implicit differentiation), tangent line equations
Linear approximation sits at the bridge between derivatives as a rate-of-change tool and derivatives as a prediction tool. If you can find a tangent line, you can estimate function values near the point of tangency, which is precisely what this section formalises. You should already be comfortable writing the equation of a tangent line given a function and a point. This topic appears regularly on the AP Calculus exam, both in multiple choice and free response.
A linear approximation uses the tangent line at a known point to estimate the value of a function at a nearby point. You are replacing the curve with its tangent line over a small interval. The closer the target x-value is to the known point, the better the estimate.
Linear approximation
The estimate of a function's value obtained by using the tangent line at a nearby point instead of the function itself.
In simple terms, this means: you swap the curve for a straight line that just touches it, then read off the straight line's y-value.
Linearization
The function L(x) = f(a) + f'(a)(x - a), which gives the tangent line to f at x = a expressed as a stand-alone function.
Think of it as: a new, simpler function that behaves like f(x) in a small neighbourhood around a.
Local linearity
The property that a differentiable function, when zoomed in sufficiently at a point, looks indistinguishable from a straight line.
In simple terms, this means: if you zoom in far enough on any smooth curve, it looks flat.
Tangent line approximation
Another name for linear approximation. The tangent line at a known point serves as the approximating tool.
Think of it as: the same concept, different label. Textbooks and the AP exam use both interchangeably.
Differentiable
A function is differentiable at a point if its derivative exists there (no corners, cusps, vertical tangents, or discontinuities).
In simple terms, this means: the function is smooth enough at that point for a tangent line to exist, which is the prerequisite for any linear approximation.
Near the point of tangency, a curve and its tangent line have nearly the same y-values.
The tangent line is easy to evaluate (it is linear), so you use it as a stand-in for the function.
The further you move from the point of tangency, the worse the approximation becomes.
Identify the base point a. Choose a value of x = a where you know (or can compute) both f(a) and f'(a). This should be close to the x-value you want to estimate.
Compute f(a). Evaluate the function at your base point.
Compute f'(a). Take the derivative and evaluate it at the base point. This gives the slope of the tangent line.
Write the linearization. Plug into L(x) = f(a) + f'(a)(x - a).
Substitute the target x-value. Replace x with the value you want to approximate.
Approximate f(-1.1) for f(x) = x^4 + 3x^2 at a = -1.
f(-1) = (-1)^4 + 3(-1)^2 = 1 + 3 = 4
f'(x) = 4x^3 + 6x, so f'(-1) = 4(-1)^3 + 6(-1) = -4 - 6 = -10
L(x) = 4 + (-10)(x - (-1)) = 4 - 10(x + 1)
L(-1.1) = 4 - 10(-1.1 + 1) = 4 - 10(-0.1) = 4 + 1 = 5
The linear approximation gives f(-1.1) approximately equal to 5.
If the function is concave up (f''(a) > 0) near the base point, the tangent line lies below the curve. The linear approximation is an underestimate.
If the function is concave down (f''(a) < 0) near the base point, the tangent line lies above the curve. The linear approximation is an overestimate.
This is a frequent AP exam question: they will ask whether your approximation is too high or too low, and you need to check concavity.
Sometimes you are not given a formula for f, only selected values of f(x) and f'(x) in a table. The process is the same: pick the table entry closest to your target x as the base point, read off f(a) and f'(a), and apply L(x) = f(a) + f'(a)(x - a).
For an implicitly defined curve (e.g. x^2 + y^3 = 2x^2 y), you find dy/dx via implicit differentiation, evaluate it at the given point, then build the tangent line equation as usual. The tangent line then serves as the linear approximation for y near that point.
The Linearization Formula (Tangent Line Approximation)
L(x) = f(a) + f'(a)(x - a)This is the equation of the tangent line to f at x = a, rewritten as a function. When you substitute any x near a, L(x) gives the approximate value of f(x).
Equivalent form using differentials:
f(x) \approx f(a) + f'(a) \, dx \quad \text{where } dx = x - aFor implicit curves where dy/dx is known at a point (a, b):
y - b = \frac{dy}{dx}\bigg|_{(a,b)} (x - a)Engineers use linear approximation constantly when exact computation is expensive or unnecessary. If a sensor reading is 3.02 and your calibration function is messy, you linearize around 3 and get a fast, good-enough answer. Physicists do the same: the small-angle approximation sin(theta) approximately equal to theta is a linear approximation around theta = 0, and it underpins everything from pendulum motion to optics.
Students often confuse the base point a with the target x-value. The base point is where you know f and f'; the target is the nearby value you are estimating. Mixing these up gives a wrong slope or a wrong starting y-value.
Students sometimes forget to evaluate the derivative at x = a and instead leave f'(x) as a general expression. The linearization requires a specific numerical slope.
A common error on implicit differentiation problems is substituting the point into the original equation instead of into dy/dx. You need dy/dx evaluated at the point to build the tangent line.
Students often assume the linear approximation is exact. It is an estimate, and it gets worse the further x moves from a. The AP exam will ask you to recognise this limitation.
This is AP Calculus AB/BC Topic 4.6. It appears in both multiple choice and free response.
The most common MC format gives you f(a) and f'(a), then asks for the estimate at a nearby x. Straight substitution into L(x) = f(a) + f'(a)(x - a). Know this cold.
Free-response questions often pair linear approximation with implicit differentiation: find dy/dx, write the tangent line at a given point, then use it to approximate y at a nearby x.
Expect a follow-up asking whether your approximation is an overestimate or underestimate. Check concavity (second derivative or the shape of the graph).
Table-based questions give discrete values of f and f' and ask you to approximate f at a value not in the table. Same formula, different packaging.
True or False: The linearization L(x) = f(a) + f'(a)(x - a) is the equation of the tangent line at x = a. True.
Fill in the blank: If f is concave up near the base point, the linear approximation is an ______. Underestimate.
True or False: You can use linear approximation even if you do not know a formula for f, as long as you know f(a) and f'(a). True.
Fill in the blank: The closer x is to a, the ______ the linear approximation. Better (more accurate).
True or False: The linear approximation of f(x) = x^2 at a = 3 gives the exact value of f(3). True. (At the base point itself, L(a) = f(a) exactly.)
Q: A differentiable function f has f(5) = 3 and f'(5) = 4. What is the estimate for f(4.8) using the local linear approximation at x = 5?
A: L(4.8) = 3 + 4(4.8 - 5) = 3 + 4(-0.2) = 3 - 0.8 = 2.2. The answer is 2.2.
Q: Find the linearization of f(x) = sqrt(x + 3) at x = 1, and use it to approximate sqrt(4.05).
A: f(1) = sqrt(4) = 2. f'(x) = 1/(2 sqrt(x + 3)), so f'(1) = 1/(2 times 2) = 1/4. L(x) = 2 + (1/4)(x - 1). To approximate sqrt(4.05), note that x + 3 = 4.05 gives x = 1.05. L(1.05) = 2 + (1/4)(0.05) = 2 + 0.0125 = 2.0125. Since f is concave down (f'' < 0), this is an overestimate.
Q: Given f'(x) = 6x^4 / (x^2 + 1) and f(1) = 8, use a linear approximation to estimate f(1.02).
A: f'(1) = 6(1)/(1 + 1) = 6/2 = 3. L(1.02) = 8 + 3(1.02 - 1) = 8 + 3(0.02) = 8.06.
Q: The local linear approximation to g at x = -3 is y = 4x - 7. What is g(-3) + g'(-3)?
A: The tangent line at x = -3 gives g(-3) = 4(-3) - 7 = -12 - 7 = -19. The slope of the tangent line is g'(-3) = 4. So g(-3) + g'(-3) = -19 + 4 = -15.
Q: For the implicit curve -8x^2 + 5xy + y^3 = -149 with dy/dx = (16x - 5y)/(5x + 3y^2), write the tangent line at (4, -1) and approximate the y-value when x = 4.2.
A: dy/dx at (4, -1) = (64 - (-5))/(20 + 3) = 69/23 = 3. The tangent line is y - (-1) = 3(x - 4), i.e. y = 3x - 13. At x = 4.2: y = 3(4.2) - 13 = 12.6 - 13 = -0.4. So k is approximately -0.4.
Linear approximation connects directly to differentials (Section 4.4), which formalise the "dx" and "dy" you see in the approximation formula. It also lays the groundwork for Taylor polynomials and Taylor series in Calculus II, where you extend the idea from a first-degree approximation to higher-degree polynomials for better accuracy. The overestimate/underestimate analysis ties back to concavity and the second derivative test from Section 4.2.
Linear approximation, linearization, tangent line approximation, local linearity, local linearization, differentials, tangent line estimate, first-order approximation, AP Calculus AB Topic 4.6, AP Calculus BC, approximating function values, overestimate vs underestimate, concavity and tangent lines, implicit differentiation tangent line, MSU Calculus I Section 4.3