From a Linear Estimate to a Diagnosis
A linear estimate is useful only if you can judge what it tells you and how much confidence to place in it. The tangent line gives a nearby, simpler value, but its accuracy depends on the function’s curvature and on how far the target lies from the base point.
This workshop uses radical, trigonometric, and exponential functions to practise a complete check: calculate the estimate, compare it with the actual value, interpret the signed error, and decide whether another base point or a shorter step would help.
The sign matters. If \(R>0\), the estimate is below the actual value; if \(R<0\), it is above. An estimate can be very close and still have a meaningful direction of error. In the examples, intermediate values are retained through subtraction so rounding does not alter the reported error.
A Reliable Diagnostic Routine
For each estimate, record the base point \(a\), the target \(x\), and the step \(h=x-a\). Then calculate \(L_a(x)\) and compare it with \(f(x)\). If a calculator provides the actual value, keep sufficient digits in both values before subtracting; subtracting rounded display values can change the final reported digit.
Substitute the base point into \(f(a)+f'(a)(x-a)\), keeping the base point and target distinct.
Use \(x-a\) for the step, with angle inputs in radians when differentiating trigonometric functions.
Calculate \(R=f(x)-L_a(x)\) before rounding, then report its sign and size.
Use concavity or a second-derivative bound to see whether the error’s direction and size are plausible.
Why Curvature Controls the Error
The tangent line matches the function’s value and slope at \(a\). Any remaining discrepancy comes from how the slope changes between \(a\) and \(x\). The following integral identity makes that idea precise for \(x\ge a\), assuming \(f\) has a continuous second derivative on \([a,x]\).
To justify the bound, the Fundamental Theorem of Calculus gives \(f(x)-f(a)=\int_a^x f'(s)\,ds\). Also, \(f'(s)-f'(a)=\int_a^s f''(t)\,dt\). Subtracting the linear term \(f'(a)(x-a)\) leaves the error:
The last equality follows by reversing the order of integration over \(a\le t\le s\le x\): for each fixed \(t\), the variable \(s\) ranges from \(t\) to \(x\). Taking absolute values and using \(|f''(t)|\le M\) yields
This result explains two practical features. First, if \(f''\) keeps one sign, the tangent-line error has a predictable direction. Second, for a fixed curvature bound, halving the step reduces the bound by a factor of four because the step is squared. The bound is a guarantee, not necessarily the exact error.
Worked Example: Estimating a Square Root
Worked Example: Estimating \(\sqrt{4.1}\)
Use the linearization of \(f(x)=\sqrt{x}\) at \(a=4\) to estimate \(\sqrt{4.1}\), then compare with the actual value and check the error bound.
First find the derivative and the values at the base point:
Thus the linearization is \(L_4(x)=2+\frac14(x-4)\). At the target \(x=4.1\), the step is \(h=4.1-4=0.1\), so
The actual value is \(\sqrt{4.1}\approx 2.024845673132\). Therefore the signed error, actual minus estimate, is
The estimate is high by about \(0.000154327\). This direction is expected because \(f''(x)=-\frac{1}{4x^{3/2}}<0\) for \(x>0\), so the graph is concave down and the tangent line lies above the function locally. For a bound on \([4,4.1]\), \(|f''(x)|\le \frac{1}{32}\), its value at \(4\). Hence
The actual absolute error, approximately \(0.000154327\), is below this bound. The check confirms both the predicted direction and the permitted size.
Worked Example: Estimating a Trigonometric Value
Worked Example: Estimating \(\sin(0.2)\)
Use \(f(x)=\sin x\) and the base point \(a=0.1\) radians to estimate \(\sin(0.2)\). Compare this with using the base point \(a=0\).
Since \(f'(x)=\cos x\), the linearization at \(0.1\) is \(L_{0.1}(x)=\sin(0.1)+\cos(0.1)(x-0.1)\). At \(x=0.2\), the step is \(0.1\). Using \(\sin(0.1)\approx0.099833416647\) and \(\cos(0.1)\approx0.995004165278\),
The actual value is \(\sin(0.2)\approx0.198669330795\). Subtracting with the unrounded values gives
The estimate is high. On \([0.1,0.2]\), \(f''(x)=-\sin x<0\), so the function is concave down and a tangent estimate from the interval’s base point is expected to lie above the function. For comparison, the base point \(a=0\) gives \(L_0(x)=x\), so \(L_0(0.2)=0.2\). Its error is \(0.198669330795-0.2=-0.001330669205\). Here the base point \(0.1\), closer to the target, gives the smaller absolute error.
The bound also provides a check. On \([0.1,0.2]\), \(|f''(x)|=\sin x\le\sin(0.2)\approx0.198669331\). Thus the error is at most \(\frac12(0.198669331)(0.1)^2\approx0.000993347\), which exceeds the actual error \(0.000664502\), as it should.
Worked Example: Estimating an Exponential Value
Worked Example: Estimating \(e^{0.2}\) and Changing the Step
Let \(f(x)=e^x\). Compare estimates of \(e^{0.2}\) made at \(a=0\) and \(a=0.1\), then examine what happens when the step from \(0.1\) increases.
Because \(f'(x)=e^x\), the linearization is \(L_a(x)=e^a+e^a(x-a)\). At \(a=0\), \(L_0(x)=1+x\), giving \(L_0(0.2)=1.2\). Since \(e^{0.2}\approx1.221402758160\), the signed error is
At \(a=0.1\), the step to \(0.2\) is \(0.1\), and
Keep full precision through the subtraction: \(e^{0.2}-1.1e^{0.1}\approx0.005715553573\), which rounds to \(0.005715554\) at nine decimal places. This positive error means the estimate is low. Since \(f''(x)=e^x>0\), the function is concave up, matching that direction. Moving the base point from \(0\) to \(0.1\) reduces the absolute error for this target from about \(0.021402758\) to \(0.005715554\).
Now keep the base point at \(0.1\) but increase the target to \(0.3\), making the step \(0.2\). The estimate is
The actual value is \(e^{0.3}\approx1.349858807576\), so the signed error is \(1.349858807576-1.326204223186\approx0.023654584390\). The larger step has produced a larger error. On \([0.1,0.3]\), \(f''(x)\le e^{0.3}\approx1.349858808\), so the bound is \(\frac12(1.349858808)(0.2)^2\approx0.026997176\), above the actual absolute error.
What to Change When an Estimate Is Unsatisfactory
If the error is too large for the task, first consider whether a closer base point is available. For a fixed target, a nearer base point often reduces the step, and the curvature bound shows why a shorter step can substantially improve accuracy. But closeness alone is not a guarantee: the curvature may vary, and the bound depends on the largest \(|f''|\) across the whole interval from base point to target.
Also use the sign as a diagnostic, not just the magnitude. A concave-up function has \(f''>0\) on the interval, so its tangent line underestimates the function there; a concave-down function has \(f''<0\), so its tangent line overestimates. If the computed sign conflicts with this prediction, check the subtraction order, derivative, base point, and step.
Rounding too early is another source of false discrepancies. For example, in the exponential comparison, subtracting displayed values that have each been rounded to only a few decimal places may not preserve the correct final digit of the error. Keep extra digits through the subtraction, then round the error once at the end.
Check Your Understanding
For each question, identify the estimate, the direction of its error, or the relevant comparison before calculating further.
- For \(f(x)=\sqrt{x}\), write the linearization at \(a=9\), then use it to estimate \(\sqrt{9.2}\).
- If \(R=f(x)-L_a(x)\) is negative, is the estimate above or below the actual value?
- For a concave-up function on the interval between a base point and target, what direction of error should its tangent-line estimate have?
- For a fixed curvature bound, how does the error bound change if the step size is cut in half?
- Why should the error be calculated before rounding the actual value and estimate?
