Introduction
A tangent line captures the value and slope of a function at one point. Nearby, it often gives a useful estimate, but it is not usually exact. The difference between the function and its tangent line is the linearization error.
The second derivative describes how the slope changes. This change in slope is what makes the graph depart from its tangent line. Near a base point, the second derivative therefore predicts the leading size of the error: for a small displacement \(h\), the error is approximately one-half the second derivative at the base point times \(h^2\).
Signed Error and the Tangent-Line Model
For a function \(f\) differentiable at \(a\), its tangent-line model at \(a\) is
Write the displacement from the base point as \(h=x-a\). The signed error is the actual function value minus the linear estimate:
This convention matters: a positive error means the function value is greater than the tangent-line estimate, and a negative error means it is less. In this tutorial, the main question is how large the error is as \(h\) becomes small, rather than a general rule for its sign.
The first-order linearization accounts for the value and slope at \(a\). What remains is the effect of changing slope. If that change is modest over a short interval, the resulting error is often much smaller than the first-order change in the function itself.
Why the Error Has a Quadratic Leading Term
Here is a precise local result. Suppose \(f''\) exists and is continuous on an interval containing \(a\). Then, as \(h\) approaches zero, the error divided by \(h^2\) approaches \(f''(a)/2\). Equivalently, the error has the form \(\frac12 f''(a)h^2\) plus a remainder that becomes negligible compared with \(h^2\).
This limit says more than “the error goes to zero.” It identifies the leading scale of the error. For a nearby input, a useful estimate is
The approximation is local: it describes behavior as \(h\) approaches zero. It is not, by itself, a guarantee that the estimate remains accurate for a large displacement.
Derivation of the Local Error Formula
We derive the result to see why both the second derivative and the squared displacement appear. Assume \(f''\) is continuous near \(a\), and let \(h\) be small enough that the interval from \(a\) to \(a+h\) lies in that neighborhood. By the Fundamental Theorem of Calculus, the error can be written as
Because \(f''\) is continuous at \(a\), the derivative \(f'\) has the local expansion
Substitute this expression into the integral. The first term integrates to a quadratic term; the remaining integral is negligible relative to \(h^2\):
To check the size of the remainder, let \(\varepsilon>0\). Since \(r(s)\to 0\), for sufficiently small \(|h|\) we have \(|r(s)|<\varepsilon\) whenever \(s\) lies between \(0\) and \(h\). Therefore
Since \(\varepsilon\) can be made arbitrarily small, the remainder divided by \(h^2\) approaches zero. Dividing the expression for \(E_a(h)\) by \(h^2\) and taking the limit gives the stated result.
Worked Example: A Quadratic Function
Worked Example: A Quadratic Function
Let \(f(x)=x^2\), with base point \(a=3\). Find the tangent-line estimate and the error at \(x=3.5\), then compare the result with the second-derivative prediction.
First, calculate the value and slope at the base point:
Thus the tangent-line model is
At \(x=3.5\), the displacement is \(h=3.5-3=0.5\). The estimate is
The actual value is \(f(3.5)=(3.5)^2=12.25\), so the signed error is \(12.25-12=0.25\). Now \(f''(x)=2\), so the quadratic prediction is
Here the prediction is exact. Indeed, direct substitution gives \(f(3+h)=9+6h+h^2\), and subtracting the tangent line \(9+6h\) leaves exactly \(h^2\). As a check, \(L_3(3)=9=f(3)\), and \(L_3'(x)=6=f'(3)\), as required for the tangent line.
Curvature and Distance from the Base Point
The local formula separates two influences. The value \(f''(a)\) measures the rate at which the slope is changing at the base point. The factor \(h^2\) measures the displacement’s contribution to the leading error. In particular, doubling a small displacement multiplies this quadratic prediction by four, provided the local approximation remains appropriate.
The absolute size of the leading term is \(\frac12|f''(a)|h^2\). Thus, among functions with the same small displacement, a larger magnitude of second derivative generally means a larger leading tangent-line error. This is a local comparison: the second derivative can change as the input moves, and the formula does not claim that the curvature at \(a\) determines the exact error at every distant input.
Worked Example: A Square-Root Function
Worked Example: A Square-Root Function
Let \(f(x)=\sqrt{x}\) and take \(a=4\). Estimate \(f(4.4)\) with the tangent line, calculate the actual error, and compare it with the local quadratic prediction.
The value and first derivative at the base point are
Therefore
The displacement is \(h=4.4-4=0.4\), so the estimate is \(L_4(4.4)=2+\frac14(0.4)=2+0.1=2.1\). The actual value is \(\sqrt{4.4}\approx 2.0976177\), rounded to seven decimal places. Hence the signed error, rounded to seven decimal places, is
For the prediction, differentiate again: \(f''(x)=-\frac{1}{4x^{3/2}}\), so \(f''(4)=-\frac{1}{32}\). The quadratic term is
The predicted error, \(-0.0025\), is close to the actual error, \(-0.0023823\). To check the tangent line, \(L_4(4)=2=f(4)\) and \(L_4'(x)=\frac14=f'(4)\). The error can also be checked by rationalizing:
At \(h=0.4\), this expression gives approximately \(-0.0023823\), in agreement with the direct subtraction.
Worked Example: How Error Changes for an Exponential
Worked Example: How Error Changes for an Exponential
For \(f(x)=e^x\), use the base point \(a=0\) and compare the errors at \(x=0.1\) and \(x=0.2\).
Since \(f(0)=1\), \(f'(x)=e^x\), and \(f'(0)=1\), the tangent line is \(L_0(x)=1+x\). Also \(f''(x)=e^x\), so \(f''(0)=1\). The local prediction at displacement \(h\) is therefore \(h^2/2\).
At \(h=0.1\), the estimate is \(L_0(0.1)=1+0.1=1.1\), while \(e^{0.1}\approx1.1051709\). Thus the actual error is approximately \(1.1051709-1.1=0.0051709\). The quadratic prediction is \(\frac12(1)(0.1)^2=0.005\).
At \(h=0.2\), the estimate is \(L_0(0.2)=1+0.2=1.2\), while \(e^{0.2}\approx1.2214028\). Thus the actual error is approximately \(1.2214028-1.2=0.0214028\). The quadratic prediction is \(\frac12(1)(0.2)^2=0.02\).
Doubling the displacement multiplies the predicted error by four: \(0.02/0.005=4\). The actual error ratio is approximately \(0.0214028/0.0051709\approx4.139\), close to four. The tangent-line check is \(L_0(0)=1=f(0)\) and \(L_0'(x)=1=f'(0)\). The predictions become more informative as the displacement becomes smaller.
When the Second Derivative at the Base Point Is Zero
If \(f''(a)=0\), the quadratic leading term vanishes. This does not prove that the tangent-line error is exactly zero, or that it must vanish at every nearby input. It says that the error divided by \(h^2\) approaches zero; a higher-order effect may then determine the error’s size.
For example, take \(f(x)=\sin x\) at \(a=0\). Its tangent line is \(L_0(x)=x\), and \(f''(0)=0\). At \(h=0.1\), the error is \(\sin(0.1)-0.1\approx0.0998334-0.1=-0.0001666\), not zero. The vanishing quadratic prediction signals that the leading error is smaller than a nonzero multiple of \(h^2\), not that there is no error.
Common Mistake: Treating Error as Proportional to Distance
A common mistake is to assume that doubling the input displacement doubles the tangent-line error. The quadratic example makes the failure clear. For \(f(x)=x^2\) at \(a=0\), the tangent line is \(L_0(x)=0\). At \(h=0.1\), the error is \((0.1)^2=0.01\). At \(h=0.2\), it is \((0.2)^2=0.04\). The displacement doubled, but the error quadrupled.
The correction is to examine the second-order scale \(h^2\), not the first-order scale \(|h|\). The local error formula gives the leading quadratic behavior when its assumptions hold. It should not be mistaken for an exact formula for every function or for a global guarantee at large distances.
Why Curvature Matters
Tangent-line estimates are useful precisely because they replace a changing function with a simple linear model near a chosen point. The second derivative tells us how rapidly the slope is changing there, and therefore how strongly the graph begins to separate from that line. This helps explain why two functions with equal values and slopes at a point can still have noticeably different linearization errors: their second derivatives may differ.
For practical use, identify the base point, compute the tangent line, and evaluate the second derivative there. The term \(\frac12 f''(a)h^2\) then gives a local guide to error size. For a more formal guarantee that an error stays within a specified amount, a bound on the second derivative is needed; that is a distinct question from the local comparison developed here.
Check Your Understanding
Try these questions using the definition of signed error and the local quadratic formula.
- For a function with \(f''(a)=6\), what is the leading error prediction at displacement \(h=0.1\)?
- If the displacement is doubled while the same local quadratic approximation applies, by what factor does the predicted error change?
- For \(f(x)=x^2\) at \(a=0\), calculate the tangent-line error at \(h=0.3\) and compare it with the second-derivative prediction.
- What does \(f''(a)=0\) imply about the quadratic term, and what does it not imply about the actual error?
- State the continuity condition used for the local error limit in this tutorial.
