4.6 Approximating Values of a Function Using Local Linearity and Linearization

Linearization uses the tangent line at a known point to estimate a nearby function value when finding the exact value is hard. Build the tangent line with the point and the derivative, plug in your target x, and solve for y. For AP Calculus, use concavity to explain whether the tangent-line estimate is too high or too low.

Why This Matters for the AP Calculus Exam

Linear approximation shows up in both multiple-choice and free-response work in AP Calculus. You may be asked to write a tangent line equation, use it to approximate a value, and then explain whether that approximation overestimates or underestimates the true value. The reasoning part matters as much as the number, since you need to connect concavity (the sign of the second derivative) to the position of the tangent line. This skill also reinforces a core idea: a differentiable function looks like a straight line when you zoom in close enough.

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Key Takeaways

• The tangent line at a point is a local linear model of the function near that point.
• You can build the tangent line with point-slope form: $y - y_1 = m(x - x_1)$, where $m = f'(a)$.
• The linearization formula $L(x) = f(a) + f'(a)(x - a)$ is the same idea written a different way.
• Plug your target x-value into the tangent line to get the approximation.
• Concave up at the point of tangency means the tangent line sits below the curve, so the estimate is an underestimate.
• Concave down at the point of tangency means the tangent line sits above the curve, so the estimate is an overestimate.

Linearization and Tangent Line Approximation

At any single point on a curve, the slope of the tangent line is the function's derivative at that point. With that slope and the coordinates of the point, you can write the tangent line using point-slope form:

$y-y_1=m(x-x_1)$

You may also see the linearization formula, which gives the same line:

$L(x)=f(a)+f'(a)(x-a)$

Here $a$ is the x-value where you know the function value and the derivative. Both forms produce the same tangent line, so use whichever you remember more reliably. Point-slope form is a solid fail-safe.

Once you have the tangent line, plug in the x-value you want to approximate and solve for y. The result is your linear approximation of the function near that point. The closer your target x-value is to $a$, the better the approximation tends to be.

Overestimates and Underestimates

A common exam task asks whether your approximation is greater than or less than the true value. You can decide this by looking at the function's concavity at the point of tangency.

If the graph is concave up at the point of tangency, the tangent line lies below the curve, so your approximation is an underestimate. Every tangent line value comes out lower than the actual function value.

If the graph is concave down at the point of tangency, the tangent line lies above the curve, so your approximation is an overestimate. Every tangent line value comes out higher than the actual function value.

A quick way to connect this to the second derivative: if $f''>0$ near the point, the function is concave up (underestimate); if $f''<0$ near the point, the function is concave down (overestimate).

How to Use This on the AP Calculus Exam

Free Response

The free-response question below is from the 2010 AP Calculus AB exam administered by College Board. All credit to College Board.

Solutions to the differential equation $\frac{dy}{dx}=xy^3$ also satisfy $\frac{d^2y}{dx^2}=y^3(1+3x^2y^2)$. Let $y=f(x)$ be a particular solution to the differential equation $\frac{dy}{dx}=xy^3$ with $f(1)=2$.

a) Write an equation for the line tangent to the graph of $y=f(x)$ at $x = 1$.

First find the slope of the tangent line at $x = 1$. Since the point $(1,2)$ is given, plug it into the derivative:

$f'(1)=(1)(2)^3=8$

Now use point-slope form to write the tangent line equation:

$y-2=8(x-1)$

b) Use the tangent line equation from part (a) to approximate $f(1.1)$. Given that $f(x)>0$ for $1<x<1.1$, is the approximation for $f(1.1)$ greater than or less than $f(1.1)$? Explain your reasoning.

Use the tangent line to estimate the function value at $x=1.1$:

$y-2=8(1.1-1)$

$f(1.1)\approx 2.8$

Because $f(x)>0$ for $1<x<1.1$, the second derivative $\frac{d^2y}{dx^2}=y^3(1+3x^2y^2)$ is positive, so the function is concave up on that interval. A concave up function sits above its tangent line, so the approximation is an underestimate of $f(1.1)$.

Common Trap

Show your reasoning for the over/underestimate claim, not just the word. Connect the concavity (or sign of the second derivative) to the position of the tangent line relative to the curve. A stated answer without that justification usually does not support a stronger score.

Common Misconceptions

• The tangent line value is not the exact function value. It is an approximation that works best very close to the point of tangency and gets worse as you move farther away.
• Concave up does not mean overestimate. Concave up puts the tangent line below the curve, which makes the approximation an underestimate. It is easy to flip these, so picture the curve bending away from the line.
• You need to check concavity at or near the point, not the slope. The slope tells you the tangent line direction, but the second derivative (concavity) decides over or underestimate.
• The two formulas are not different methods. $L(x)=f(a)+f'(a)(x-a)$ and point-slope form describe the same tangent line. Pick one and use it consistently.
• "Locally linear" means nearby, not everywhere. A linear approximation is only reliable in a small region around the point of tangency.

Related AP Calculus Guides

• Unit 4 Overview: Contextual Applications of Differentiation
• 4.1 Interpreting the Meaning of the Derivative in Context
• 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
• 4.4 Intro to Related Rates
• 4.5 Solving Related Rates Problems
• 4.3 Rates of Change in Applied Contexts other than Motion