When Is a Tangent-Line Estimate Credible?
A tangent line gives a local model for a function near a chosen base point. But a correct tangent line is not automatically a useful estimate at every input. The question is whether the model is accurate enough for the particular input and purpose.
To assess that, check how far the target input lies from the base point, how much the function’s slope changes over that interval, whether the function is defined throughout it, and how much error the situation allows. These checks work together: no single rule, such as “a small input change is safe,” is enough in every context.
Four Checks for a Local Model
Start by comparing the target \(x\) with the base point \(a\). The input displacement is \(\Delta x=x-a\), and the model uses the function’s value and slope at \(a\) to predict the change. A smaller displacement is often helpful, but “small” is relative: a change of \(0.1\) may be tiny for one quantity and substantial for another.
Next, consider curvature: how quickly the slope changes. If \(f''\) is small throughout the interval, the slope stays closer to its value at the base point, so the tangent line is more likely to track the function well. When \(f''\) is large, the slope can change quickly and the tangent-line estimate may become less accurate, even over a modest input interval.
Also check the function’s domain. The function must be defined at the target, and its behavior between the base point and target must support the local model. A tangent-line formula can produce a number even when the original function has no value at the target; that number is not then an approximation to a defined function value.
Finally, consider how rapidly the function changes in output. A large value of \(|f'(a)|\) means that a small input change can produce a large output change according to the tangent line. This matters when interpreting the estimate, but it does not by itself mean the estimate is inaccurate. Accuracy depends on how much the slope changes, not merely on the size of the slope at \(a\).
Using a Curvature Bound
A second-derivative bound can turn the curvature check into a quantitative test. Suppose \(f''\) is continuous on the interval connecting \(a\) and \(x\), and suppose \(|f''(t)|\leq M\) throughout that interval. Then the linearization error is bounded by \(M|x-a|^2/2\). This is a guarantee on the possible error, not necessarily the exact error.
The bound helps compare the model with a required tolerance. If the bound is below the tolerance, the model is guaranteed to meet it. If the bound is above the tolerance, this bound alone does not decide the question: the actual error might still be smaller.
Why the Curvature Bound Works
The tangent line matches both the function’s value and its slope at \(a\). The error accumulates from the difference between the function’s slope at an intermediate input and its slope at \(a\). The second derivative controls how large that slope difference can be.
For \(x\geq a\), the Fundamental Theorem of Calculus gives $$ f(x)-L_a(x)=\int_a^x\bigl(f'(t)-f'(a)\bigr)\,dt, $$ because \(L_a(x)=f(a)+f'(a)(x-a)\).
For \(a\leq t\leq x\), the same theorem gives \(f'(t)-f'(a)=\int_a^t f''(u)\,du\). Thus $$ |f'(t)-f'(a)|\leq\int_a^t |f''(u)|\,du\leq M(t-a). $$
Therefore $$ |f(x)-L_a(x)|\leq\int_a^x M(t-a)\,dt =\frac{M}{2}(x-a)^2. $$ For \(x<a\), reversing the interval gives the same bound with \(|x-a|\). This proves the result on either side of the base point.
The squared distance is important: if the displacement is halved while the same curvature bound applies, the error bound becomes one quarter as large. But the value of \(M\) also matters. Near a point where curvature grows rapidly, a short distance may not make the error acceptably small.
Worked Example: A Nearby Estimate with a Small Error Bound
Worked Example: Estimating a Square Root Near 4
Use the local linear model for \(f(x)=\sqrt{x}\) at \(a=4\) to estimate \(\sqrt{4.2}\). Then decide whether the error is guaranteed to be less than \(0.001\).
First, \(f(4)=2\), and \(f'(x)=1/(2\sqrt{x})\), so \(f'(4)=1/4\). The model is $$ L_4(x)=2+\frac14(x-4). $$
At \(x=4.2\), the displacement is \(4.2-4=0.2\), and $$ L_4(4.2)=2+\frac14(0.2)=2+0.05=2.05. $$
For the error bound, \(f''(x)=-1/(4x^{3/2})\). On \([4,4.2]\), \(|f''(x)|\leq 1/32\), since \(4^{3/2}=8\) and \(1/(4\cdot 8)=1/32\). Thus $$ |f(4.2)-L_4(4.2)|\leq \frac12\left(\frac1{32}\right)(0.2)^2 =\frac{0.04}{64}=0.000625. $$ This is below \(0.001\), so the estimate meets the stated tolerance. As a check, \(\sqrt{4.2}\approx 2.049390\), which differs from \(2.05\) by about \(0.000610\), consistent with the bound.
Worked Example: Checking Whether the Tolerance Is Met
Worked Example: Estimating a Reciprocal Away from the Base Point
Use \(f(x)=1/x\) at \(a=2\) to estimate \(f(2.5)\). Check whether the estimate is within \(0.01\) of the actual value.
We have \(f(2)=1/2\), \(f'(x)=-1/x^2\), and \(f'(2)=-1/4\). Therefore $$ L_2(x)=\frac12-\frac14(x-2), \qquad L_2(2.5)=\frac12-\frac14(0.5)=0.5-0.125=0.375. $$
The interval \([2,2.5]\) lies in the domain \(x\neq 0\). Also, \(f''(x)=2/x^3\), so \(|f''(x)|\leq 1/4\) on this interval. The error bound is $$ |f(2.5)-L_2(2.5)|\leq \frac12\left(\frac14\right)(0.5)^2 =\frac18(0.25)=0.03125. $$
This bound is larger than \(0.01\), so it does not guarantee the required accuracy. In fact, \(f(2.5)=1/2.5=0.4\), and the actual error is $$ |0.4-0.375|=0.025. $$ Thus the estimate does not meet the tolerance. Checking the actual value confirms the conclusion; the bound alone would not have been enough to reject the estimate.
Worked Example: A Valid Input Near a Domain Boundary
Worked Example: Estimating a Logarithm Near Zero
Use \(f(x)=\ln x\) at \(a=1\) to estimate \(\ln(0.1)\). Examine whether the estimate is likely to be reliable.
Since \(f(1)=0\) and \(f'(x)=1/x\), we have \(f'(1)=1\). The local model gives $$ L_1(x)=0+1(x-1)=x-1, \qquad L_1(0.1)=0.1-1=-0.9. $$
The target is in the domain, and the whole interval \([0.1,1]\) is within the domain \(x>0\). However, \(f''(x)=-1/x^2\), so \(|f''(x)|\leq 100\) on this interval. The bound is $$ |f(0.1)-L_1(0.1)|\leq \frac{100}{2}(0.1-1)^2 =50(0.81)=40.5. $$
This very large bound does not establish a useful level of accuracy. Indeed, \(\ln(0.1)\approx -2.302585\), so the actual error is $$ |-2.302585-(-0.9)|\approx 1.402585. $$ The interval stays in the domain, but it approaches the boundary \(x=0\), where the magnitude of the derivative grows. The tangent line at \(1\) is therefore a poor estimate at \(0.1\). The numerical comparison confirms this diagnosis.
Common Pitfall: Treating the Tangent Formula as a Domain Extension
A tangent-line formula is defined algebraically wherever its expression makes sense. That does not mean it gives an approximation to the original function outside the function’s domain.
For example, with \(f(x)=\ln x\) at \(a=1\), the tangent line is \(L_1(x)=x-1\). Substituting \(x=-0.1\) gives \(L_1(-0.1)=-1.1\). The incorrect conclusion would be that \(\ln(-0.1)\approx -1.1\). But \(\ln(-0.1)\) is not defined as a real number, because the real logarithm requires a positive input. The correction is to check the domain before interpreting the estimate: there is no real function value at that target for the tangent line to approximate.
Rapid Change and Approximation Accuracy
A large slope can indicate that output values are sensitive to input changes. That can be important when deciding whether a small measurement uncertainty is acceptable. But slope size and linearization error answer different questions: the slope describes the model’s rate of change at the base point, while curvature describes how the rate of change varies nearby.
For instance, if \(f(x)=100x\), then \(f'(x)=100\), so a change of \(0.01\) in \(x\) corresponds to a change of \(1\) in the model’s output. Yet the function is itself linear, so its tangent-line model is exact. A large output change is not the same as a large approximation error.
In practice, decide what accuracy is needed first. Then verify the domain and interval, consider the displacement and how it compares with the function’s local behavior, and use a curvature bound when available. If the bound is too large to settle the question, use a closer base point, a smaller step, or a direct comparison when the function value can be evaluated.
Check Your Understanding
For each question, explain which check matters and what conclusion it supports.
- Why does a small input displacement not guarantee a good linear estimate on its own?
- If a second-derivative bound gives an error bound below the required tolerance, what does that guarantee?
- If the error bound is above the tolerance, does that prove the estimate fails? Explain.
- Why can a function with a large derivative still have an exact tangent-line model?
- What domain check is needed before using a tangent-line estimate at a target input?
