Why the Direction of an Estimate Matters
A tangent line gives a local linear estimate of a function. Knowing whether that estimate is likely to be above or below the actual function value adds useful information: it tells us the direction of the error, not just that an error may exist.
The key is the function’s concavity over the interval between the base point and the point being estimated. If the graph bends upward throughout that interval, its tangent line lies below the graph. If it bends downward throughout, the tangent line lies above. The condition must hold across the interval, not merely at the base point.
Concavity and the Tangent Line
For a twice-differentiable function, the sign of \(f''\) describes how the slopes of the graph change. When \(f''(x)>0\) throughout an interval, the slopes increase and the graph is concave up there. When \(f''(x)<0\) throughout an interval, the slopes decrease and the graph is concave down. Non-strict inequalities allow the second derivative to be zero at some points.
These conclusions concern the whole segment between \(a\) and \(x\), regardless of whether \(x\) is to the left or right of \(a\). If the curvature changes sign along that segment, this theorem does not determine the error sign.
Why the Sign Rule Works
The second-order remainder formula makes the connection exact: for \(x\ne a\), the error equals one-half the second derivative at some intermediate point, multiplied by the square of the distance from the base point. Here is a derivation using Rolle’s theorem.
Let \(E=f(x)-L_a(x)\), and let \(K=E/(x-a)^2\). Define \(q(t)=f(t)-L_a(t)-K(t-a)^2\) on the segment from \(a\) to \(x\).
By the definition of \(L_a\), \(q(a)=0\). By the definition of \(K\), \(q(x)=E-K(x-a)^2=0\). Also, \(q'(a)=f'(a)-f'(a)-2K(a-a)=0\).
Since \(q(a)=q(x)\), Rolle’s theorem gives a point \(r\) strictly between \(a\) and \(x\) with \(q'(r)=0\). Since \(q'(a)=q'(r)=0\), a second application gives a point \(c\) strictly between \(a\) and \(r\) with \(q''(c)=0\).
Because \(L_a''(t)=0\), we have \(q''(t)=f''(t)-2K\). Therefore \(q''(c)=0\) implies \(K=f''(c)/2\), and hence the exact error formula follows.
The square \((x-a)^2\) is positive whenever \(x\ne a\). Therefore the sign of the error is the sign of \(f''(c)\). If \(f''\) is nonnegative throughout the segment, then \(f''(c)\geq 0\) and the error is nonnegative. If \(f''\) is nonpositive throughout, the error is nonpositive. This proves the sign rule. If the second derivative is strictly positive or strictly negative throughout the segment, the error is strictly positive or strictly negative, respectively.
Worked Examples
Worked Example: Concave Up and an Underestimate
Use the linearization of \(f(x)=x^2\) at \(a=1\) to estimate \(f(7/5)\). Determine the direction of the error and check it exactly.
First, \(f(1)=1\) and \(f'(x)=2x\), so \(f'(1)=2\). The linearization is
At \(x=7/5\),
Also, \(f''(x)=2>0\) throughout the segment from \(1\) to \(7/5\). The theorem predicts an underestimate. Check the actual value and error:
Thus the linear estimate \(9/5\) is below the exact value \(49/25\), as predicted.
Worked Example: Concave Down and an Overestimate
Use the linearization of \(f(x)=-x^2+4x+1\) at \(a=1\) to estimate \(f(3/2)\). Determine the direction of the error and check it exactly.
At the base point,
Therefore,
Since \(f''(x)=-2<0\) throughout the segment from \(1\) to \(3/2\), the linearization should overestimate. Direct calculation verifies this:
The estimate \(5\) is above the exact value \(19/4\), as predicted.
Worked Example: Curvature Changes Between the Points
Consider \(f(x)=x^3\), with base point \(a=-1\) and target \(x=3\). Find the linearization error and explain why the curvature at the base point alone cannot predict its sign.
At the base point,
The linearization and its value at the target are
The actual function value is \(f(3)=3^3=27\), so
In this case, the linearization underestimates. But \(f''(x)=6x\), so \(f''(-1)=-6<0\): the function is concave down at the base point. That fact alone would not justify predicting an overestimate at \(x=3\). The second derivative changes sign at \(0\) within the segment from \(-1\) to \(3\), so neither constant-sign condition in the theorem holds throughout the segment. The actual error must be checked by another method.
A Common Mistake
A frequent error is to reverse the connection between concavity and tangent-line position. For example, someone might reason, “\(f(x)=x^2\) is concave up, so its tangent line overestimates.” At \(a=1\), that would lead to the false claim that \(L_1(7/5)=9/5\) is above \(f(7/5)=49/25\). But \(9/5=45/25<49/25\), so the linearization is an underestimate.
When This Sign Test Is Useful
In a contextual problem, a linearization provides an estimate for a quantity that changes with an input. Concavity tells whether that estimate tends to be high or low, which can guide interpretation of the result. For instance, if a model is concave up throughout the relevant interval, the tangent-line estimate is below the modeled value; if it is concave down throughout, the estimate is above it.
This is a directional conclusion, not a measurement of how large the error is. A sign test tells whether the estimate over- or underestimates, but it does not by itself give the error’s magnitude. It also does not apply merely because the function has a particular concavity at the base point: the curvature condition must hold across the interval to the target. When curvature changes sign, inspect the function and the tangent line directly or use another justified method.
Check Your Understanding
For each question, keep the error convention \(E(x)=f(x)-L_a(x)\) in mind.
- If \(f''(t)>0\) throughout the segment from \(a\) to \(x\), is the linearization an overestimate or an underestimate?
- If \(f''(t)<0\) throughout the segment, what is the sign of \(E(x)\)?
- Why does the sign of \(f''(a)\) alone fail to determine the error when the interval from \(a\) to \(x\) contains a change in concavity?
- For \(f(x)=x^2\) at \(a=2\), write \(L_2(x)\). Without evaluating a specific target, determine the error sign for any \(x\ne 2\).
- In the exact remainder formula, why does the factor \((x-a)^2\) not change the sign of the error?
