AP®MathsCollege BoardCalculus BCRevision NotesUnit 4: Contextual Applications of DifferentiationLinearizationApproximating Values of a Function

Approximating Values of a Function (College Board AP® Calculus BC): Revision Note

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 6 May 2026

Local linearity of a function

What does local linearity mean?

If you 'zoom in' far enough on the graph of a function at a point, a curve can look more like a straight line
This means the tangent to a graph of a function at a point, can act as an approximation for the function at that point
This linear approximation of a function is only appropriate very close to the point
- Hence the term "local linearity"

Graphs of y = x^3 - 4x + 3 (black curve) and y = 8x - 13 (red line), intersecting at point (2, 3), shown in blue, with zoomed-in view on the right. — The graph of a cubic (black) and the graph of the tangent (red) to the curve at (2,3). The view on the right is zoomed in. You can see that close to the point, the curve is approximated by the straight line.

How do I use a tangent to approximate a function?

The equation of the tangent to $f (x)$ at $x = a$ is given by
- $y - f (a) = f^{'} (a) (x - a)$
- Provided that $f (x)$ is differentiable at $a$
Due to the local linearity of a function this can be a linear approximation for $f (x)$ at points close to $(a, f (a))$
The linearization function of $f$ at $x = a$ can be written as the function
- $L (x) = f (a) + f^{'} (a) (x - a)$

Using the example in the above image
- For the graph of $y = x^{3} - 4 x + 3$
- The tangent at (2, 3) is $y = 8 x - 13$
- The tangent will be an approximation for the curve close to (2,3)
- Substitute $x$ values close to 2 into the equation of the tangent to find an approximation for the function (curve) at that point
- See the table below for the approximated values compared to the real values

$x$	$y = 8 x - 13$ (Tangent)	$y = x^{3} - 4 x + 3$ (Curve)
2.3	5.4	5.967
2.2	4.6	4.848
2.1	3.8	3.861
2.01	3.08	3.0806
2	3	3
1.99	2.92	2.9205
1.9	2.2	2.259
1.8	1.4	1.632
1.7	0.6	1.113

The table shows how the approximation is more accurate closer to the point where the tangent intersects the curve
- It will be less accurate further away from the point of intersection
Using a tangent to approximate a curve (within a small interval) can make calculations or computational processes easier to handle,
- This is because a linear function is simpler than most other functions
- However, this comes with a trade-off in accuracy

Examiner Tips and Tricks

These questions normally appear on Part B (non-calculator part) of the FRQ section alongside implicit differentiation. Make sure you show your calculation so that you can still score points if you make an arithmetic error.

Worked Example

Consider the curve defined by the equation $3 x^{2} - 2 y + y^{2} = 11$ .

(a) Show that $\frac{d y}{d x} = \frac{3 x}{1 - y}$ .

(b) There is a point on the curve near $(1, 4)$ with an $x$ -coordinate of 1.2. Use the line tangent to the curve at $(1, 4)$ to approximate the $y$ -coordinate of this point.

Answer:

(a)

Differentiate both sides of the equation with respect to $x$

$\begin{array}{rcl} \frac{d}{d x} (3 x^{2}) - \frac{d}{d x} (2 y) + \frac{d}{d x} (y^{2}) & = & 11 \\ 6 x - 2 \frac{d y}{d x} + 2 y \frac{d y}{d x} & = & 0 \end{array}$

Divide both sides by 2

$\begin{array}{rcl} 3 x - \frac{d y}{d x} + y \frac{d y}{d x} & = & 0 \end{array}$

Isolate the terms in $\frac{d y}{d x}$ on one side

$\begin{array}{rcl} 3 x & = & \frac{d y}{d x} - y \frac{d y}{d x} \end{array}$

Factor the $\frac{d y}{d x}$ from the right-hand side

$\begin{array}{rcl} 3 x & = & (1 - y) \frac{d y}{d x} \end{array}$

Divide both sides by the factor to get the required result

$\frac{d y}{d x} = \frac{3 x}{1 - y}$

(b)

Substitute $x = 1$ and $y = 4$ into $\frac{d y}{d x}$ to find the slope of the tangent

${\frac{d y}{d x}}_{x = 1, y = 4} = \frac{3 \cdot 1}{1 - 4} = \frac{3}{- 3} = - 1$

Find the equation of the linear approximation using $L (x) = f (a) + f^{'} (a) (x - a)$

$\begin{array}{rcl} L (x) & = & 4 + (- 1) \cdot (x - 1) \\ = & 4 - (x - 1) \\ = & 5 - x \end{array}$

Find the approximation for $y$ when $x = 1.2$ by substituting this value into the function

$\begin{array}{rcl} L (1.2) & = & 5 - 1.2 \\ = & 3.8 \end{array}$

When $x = 1.2$ , $y \approx 3.8$

How do I know if the approximation is an overestimate or underestimate?

The values of the function approximated by the tangent will be either an overestimate or underestimate of the real value
Which one it is depends on the concavity of the function at the point where the tangent intersects the curve
- You can find out more about concavity in the 'Concavity of Functions' study guide
In general,
- If the graph of the function is concave up $(f^{''} (x) > 0)$ at the point where the tangent intersects it, the tangent will give an underestimate
- If the graph of the function is concave down $(f^{''} (x) < 0)$ at the point where the tangent intersects it, the tangent will give an overestimate

Graph showing y=f(x), its tangents at x=-2 and x=2 in red, and its second derivative y=f''(x) in blue, with labeled axes and points. — **The graph of f(x) is concave down for negative values of x, and concave up for positive values of x. This is also shown by the graph of the second derivative of f(x).**

In the graph above:
- The tangent at $x = - 2$ will give an overestimate, as the function is concave down at this point
- The tangent at $x = 2$ will give an underestimate, as the function is concave up at this point

Worked Example

Consider the curve defined by the equation $y = \sqrt{x}$ .

(a) Use the line tangent to the curve at the point $x = 64$ to approximate the value of $\sqrt{65}$ .

(b) Without calculating the real value of $\sqrt{65}$ , explain whether your approximation will be an overestimate or underestimate.

Answer:

(a)

Find the y-value

$y = \sqrt{64} = 8$

Differentiate the curve

$\frac{d y}{d x} = \frac{1}{2} x^{- \frac{1}{2}}$

Find the slope of the tangent at $x = 64$

$\frac{d y}{d x} = \frac{1}{2} \cdot \frac{1}{\sqrt{64}} = \frac{1}{16}$

Find the function of the linear approximation using $L (x) = f (a) + f^{'} (a) (x - a)$

$\begin{array}{rcl} L (x) & = & f (64) + f^{'} (64) (x - 64) \\ = & 8 + \frac{1}{16} (x - 64) \end{array}$

Use the tangent to estimate the value of $\sqrt{65}$ by substituting in $x = 65$

$\begin{array}{rcl} L (x) & = & 8 + \frac{1}{16} (65 - 64) \\ = & 8 + \frac{1}{16} = 8.0625 \end{array}$

An approximation of $\sqrt{65}$ is 8.0625

(b)

Consider the concavity of the curve to decide if the tangent at $x = 64$ will be an over- or underestimate

$\frac{d^{2} y}{d x^{2}} = - \frac{1}{4} x^{- \frac{3}{2}}$

Note that $x^{- \frac{3}{2}} = \frac{1}{x \sqrt{x}}$ is only defined for $x > 0$ , and for those values of $x$ it is always positive

The second derivative is always negative, so the graph of $y = \sqrt{x}$ is always concave down

Therefore the approximation using a tangent will be an overestimate

You can also see this when sketching a graph of $y = \sqrt{x}$ and the tangent at (64, 8)

The tangent is always above the curve, so will be an overestimate

Graph of root x and a tangent to it at (64, 8). The tangent is always above the curve.

In fact, $\sqrt{65} = 8.0622 . . .$ which is very close to the apporximation

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Approximating Values of a Function (College Board AP® Calculus BC): Revision Note

Local linearity of a function

What does local linearity mean?

How do I use a tangent to approximate a function?

Examiner Tips and Tricks

Worked Example

How do I know if the approximation is an overestimate or underestimate?

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Jamie Wood

Reviewer: Dan Finlay