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Mean Value Theorems · Tutorial 702 of 1495

Error Bounds for Linear Approximation from the Mean Value Theorem

Learn to turn derivative information on an interval into a rigorous bound for the error of a linear approximation.

Advanced 9 min read

What You'll Learn

  • Write the linear approximation to a function at a reference input.
  • Express its error exactly using the Mean Value Theorem.
  • Bound the error when the first derivative varies by a known amount.
  • Derive a quadratic error bound from a bound on the second derivative.
  • Check the interval and differentiability conditions needed for each bound.

Why Bound a Linear Approximation?

A tangent line replaces a function locally by a simpler linear function. This can make a value easier to estimate, but an estimate is useful only if we know how far it may be from the actual function value. The Mean Value Theorem gives an exact description of that error: it is the input change multiplied by the difference between two derivative values.

When the derivative itself changes at a controlled rate, the error is often much smaller than a simple bound based only on the size of the derivative. In particular, a bound on the second derivative gives an error proportional to the square of the input change. The interval on which the derivative bound holds matters: it must include the entire segment between the reference input and the input being estimated.

Linear Approximation and Its Error

Let \(f\) be differentiable at a reference input \(a\). Its linear approximation at \(a\), evaluated at \(x\), is the value on the tangent line at \(a\).

$$ L_a(x)=f(a)+f'(a)(x-a). $$

The error is the difference between the actual value and this approximation. We use the signed error \(E_a(x)\); its absolute value is the error magnitude.

Definition: The signed error in the linear approximation of \(f(x)\) at \(a\) is $$ E_a(x)=f(x)-L_a(x)=f(x)-f(a)-f'(a)(x-a). $$

If \(x=a\), then \(E_a(a)=0\) exactly. For \(x\ne a\), the Mean Value Theorem applies if \(f\) is continuous on the closed interval with endpoints \(a\) and \(x\), and differentiable in its interior. It gives a point \(c\) strictly between \(a\) and \(x\) such that

$$ f(x)-f(a)=f'(c)(x-a). $$

Subtracting the tangent-line term gives an exact error identity.

Theorem: If \(x\ne a\), and \(f\) is continuous on the closed interval with endpoints \(a,x\) and differentiable in its interior, then some \(c\) strictly between \(a\) and \(x\) satisfies $$ E_a(x)=\bigl(f'(c)-f'(a)\bigr)(x-a). $$

This identity is exact, but the point \(c\) is generally not known. To obtain a numerical bound, we need information that controls how far \(f'(c)\) can be from \(f'(a)\).

A Bound Using First-Derivative Information

Suppose that throughout the interval between \(a\) and \(x\), the derivative differs from \(f'(a)\) by at most \(D\). The exact error identity immediately gives a bound. The condition must hold on the intervening interval, not merely at its endpoints.

Theorem: Under the hypotheses of the Mean Value Theorem above, if $$ |f'(t)-f'(a)|\le D $$ for every \(t\) between \(a\) and \(x\), where \(D\ge 0\), then $$ |E_a(x)|\le D|x-a|. $$

Indeed, the identity supplies an interior point \(c\), and the assumed inequality applies to that point. Taking absolute values gives \(|E_a(x)|=|f'(c)-f'(a)|\,|x-a|\le D|x-a|\).

A bound on the size of the derivative can also be used, although it may be less sharp. If \(|f'(t)|\le B\) throughout the interval, then \(|f'(c)-f'(a)|\le |f'(c)|+|f'(a)|\le 2B\), so \(|E_a(x)|\le 2B|x-a|\). This bound does not say that the derivative changes slowly; it only limits its size.

A Quadratic Error Bound

A stronger and often more informative result is available when the second derivative is bounded. Suppose \(f\) is twice continuously differentiable on an interval containing the closed interval with endpoints \(a,x\), and that \(|f''(t)|\le K\) there. Then the linear approximation error is at most \(K|x-a|^2/2\).

Theorem: If \(f\) is twice continuously differentiable on an interval containing \(a\) and \(x\), and $$ |f''(t)|\le K $$ throughout the interval between them, then $$ |f(x)-f(a)-f'(a)(x-a)|\le \frac{K}{2}|x-a|^2. $$

The squared input difference explains why the tangent-line approximation improves rapidly as \(x\) approaches \(a\). The factor \(1/2\) is important. A direct application of the Mean Value Theorem to the error, after bounding its derivative, gives a valid but generally weaker bound. The following argument proves the sharper factor.

Why the Quadratic Bound Holds

We prove a more precise statement: for \(x\ne a\), the error equals one half of a second-derivative value at some intermediate point, multiplied by \((x-a)^2\). This proof uses Rolle’s Theorem, which follows from the Mean Value Theorem material on repeated roots.

1
Fix the error.
Let \(h=x-a\ne 0\) and \(E=E_a(x)\). Compare \(f\) to the quadratic function $$ q(t)=f(a)+f'(a)(t-a)+\frac{E}{h^2}(t-a)^2. $$
2
Find zeros of the difference.
Set \(g(t)=f(t)-q(t)\). By the definitions of \(E\) and \(h\), \(g(a)=0\) and \(g(x)=0\). Also \(g'(a)=f'(a)-f'(a)=0\).
3
Apply Rolle’s Theorem twice.
The two endpoint zeros give a point \(u\) strictly between \(a\) and \(x\) with \(g'(u)=0\). Since \(g'(a)=g'(u)=0\), applying Rolle’s Theorem to \(g'\) between \(a\) and \(u\) gives a point \(v\) strictly between them with \(g''(v)=0\).
4
Recover the error.
Because \(q''(t)=2E/h^2\), the equality \(g''(v)=0\) says \(f''(v)=2E/h^2\). Therefore $$ E_a(x)=\frac{f''(v)}{2}(x-a)^2. $$

Taking absolute values and using \(|f''(v)|\le K\) proves the stated bound. When \(x=a\), the error is zero, so the bound holds there as well, without requiring an intermediate point.

Worked Examples

Worked Example: An Exact Error for a Quadratic

Approximate \(f(1.1)\) using the tangent line to \(f(x)=x^2\) at \(a=1\), then bound the error using a second-derivative bound.

First, \(f(1)=1\) and \(f'(x)=2x\), so \(f'(1)=2\). The linear approximation is

$$L_1(1.1)=1+2(1.1-1)=1+0.2=1.2.$$

Since \(f''(x)=2\), take \(K=2\) on the interval from \(1\) to \(1.1\). The bound is

$$|E_1(1.1)|\le \frac{2}{2}(1.1-1)^2=1(0.1)^2=0.01.$$

The exact function value is \(f(1.1)=(1.1)^2=1.21\), so the signed error is \(1.21-1.2=0.01\). Its magnitude equals the bound. As a check, the exact error identity gives \(E_1(1.1)=f''(c)(0.1)^2/2=2(0.01)/2=0.01\), consistent with the calculation.

Worked Example: Bounding an Approximation to a Reciprocal

Use the tangent line to \(f(x)=1/(1+x)\) at \(a=0\) to approximate \(f(0.1)\), and find a rigorous error bound.

We have \(f(0)=1\) and \(f'(x)=-1/(1+x)^2\), so \(f'(0)=-1\). Thus

$$L_0(0.1)=1-(0.1)=0.9.$$

The second derivative is \(f''(x)=2/(1+x)^3\). On \([0,0.1]\), \(1+x\ge1\), so \(0<f''(x)\le2\). We may use \(K=2\). Therefore

$$|E_0(0.1)|\le \frac{2}{2}(0.1)^2=0.01.$$

The exact value is \(f(0.1)=1/1.1=10/11\). The signed error is \(10/11-9/10=(100-99)/110=1/110\), which is positive. Since \(1/110<1/100=0.01\), the actual error obeys the bound. The derivative bound was checked on the full interval between \(0\) and \(0.1\), as required.

Worked Example: Estimating a Square Root Near Four

Use the tangent line to \(f(x)=\sqrt{x}\) at \(a=4\) to approximate \(\sqrt{4.1}\) and bound its error.

Since \(f(4)=2\) and \(f'(x)=1/(2\sqrt{x})\), we have \(f'(4)=1/4\). With \(x=4.1\), the linear approximation is

$$L_4(4.1)=2+\frac14(4.1-4)=2+\frac{1}{40}=\frac{81}{40}=2.025.$$

For \(x>0\), \(f''(x)=-1/(4x^{3/2})\). Its absolute value decreases as \(x\) increases, so on \([4,4.1]\) it is at most \(1/(4\cdot4^{3/2})=1/32\). Set \(K=1/32\). The error bound is

$$|E_4(4.1)|\le \frac{1/32}{2}(0.1)^2=\frac{1}{6400}=0.00015625.$$

For a direct check of the approximation’s direction and error, \((81/40)^2=6561/1600\), whereas \(4.1=6560/1600\). Thus \(81/40>\sqrt{4.1}\), and rationalizing gives

$$\frac{81}{40}-\sqrt{4.1}=\frac{(81/40)^2-4.1}{81/40+\sqrt{4.1}}=\frac{1/1600}{81/40+\sqrt{4.1}}.$$

The denominator is greater than \(4\), because \(81/40>2\) and \(\sqrt{4.1}>2\). Hence the error is less than \(1/6400\), in agreement with the theorem.

Choosing and Applying a Bound

The useful bound depends on what derivative information is available. A direct bound on \(|f'(t)-f'(a)|\) gives an error proportional to \(|x-a|\). A second-derivative bound gives a quadratic error estimate. Do not replace one hypothesis with the other: a bound on \(|f'|\) alone does not imply a small change in \(f'\) near \(a\).

  • Identify the reference input \(a\), the target input \(x\), and the whole interval between them.
  • Write \(L_a(x)=f(a)+f'(a)(x-a)\) before estimating.
  • Use a first-derivative variation bound if you know \(|f'(t)-f'(a)|\le D\).
  • Use the quadratic bound if \(f\) is twice continuously differentiable and \(|f''(t)|\le K\) throughout the interval.
  • Keep the absolute value on the error unless its sign has also been established.

Common Pitfall: Using a Derivative Bound on the Wrong Quantity

A frequent mistake is to use a bound on \(|f'(t)|\) as though it were a bound on \(|f'(t)-f'(a)|\). For the reciprocal example, \(|f'(t)|\le1\) on \([0,0.1]\), but that does not justify the claim \(|E_0(0.1)|\le 1(0.1)^2/2\). The quadratic theorem requires a bound on the second derivative, not just the first.

Here the correct second-derivative bound is \(|f''(t)|\le2\), which gives \(|E_0(0.1)|\le0.01\). Alternatively, since \(|f'(t)-f'(0)|\le1\) on the interval, the first-derivative variation result gives the valid bound \(|E_0(0.1)|\le1(0.1)=0.1\), which is weaker. The two estimates use different hypotheses and should not be interchanged.

Why the Interval and Conditions Matter

All derivative bounds must hold between the two inputs. A bound that holds only at \(a\), or only at \(x\), does not control the derivative at the intermediate point supplied by the Mean Value Theorem. Likewise, when \(x\ne a\), the theorem requires continuity on the closed interval and differentiability in its interior. The second-derivative estimate uses the stronger stated smoothness assumptions.

The result is useful whenever a complicated function is replaced by a tangent line and a guaranteed tolerance is needed. Because the error bound scales like \(|x-a|^2\) under a second-derivative bound, halving the input distance reduces the guaranteed error by a factor of four, provided the same bound \(K\) remains valid on the smaller interval.

Check Your Understanding

Answer each question by identifying the needed derivative information and checking the interval involved.

  1. Write the linear approximation to \(f(x)=\ln x\) at \(a=1\).
  2. If \(|f'(t)-f'(a)|\le 0.03\) between \(a\) and \(x\), and \(|x-a|=0.2\), what error bound follows?
  3. If \(|f''(t)|\le 6\) and \(|x-a|=0.05\), what quadratic error bound follows?
  4. Why does a bound on \(|f'(t)|\) alone not generally give the quadratic error estimate?
  5. What does the Mean Value Theorem guarantee when \(x\ne a\), and why is that condition necessary?

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