From a Function Formula to a Nearby Estimate
A function formula can give an exact value at any input where the function is defined. But evaluating some functions exactly can be awkward, even when the input is close to a value that is easy to use. Linearization provides a practical estimate: calculate the function and its slope at a convenient base point, then use the tangent line there to estimate the nearby value.
The main task in this tutorial is to use formula data correctly. For a specified function, base point, and target input, you will calculate the linearization and evaluate it at the target. The same procedure works for polynomial, rational, and radical functions as long as the function is differentiable at the base point.
The Calculation to Perform
Suppose \(f\) is differentiable at \(x=a\). The linearization of \(f\) at \(a\) is
To estimate \(f(b)\), where \(b\) is the specified input near \(a\), evaluate \(L(b)\):
The base point and target input have different jobs. The base point \(a\) determines both the function value \(f(a)\) and the slope \(f'(a)\). The target input \(b\) is substituted only after the linearization has been formed. The difference \(b-a\) is the signed change from the base point; it is negative when the target lies to the left of \(a\).
Why the Formula Gives a Tangent-Line Estimate
At the point \((a,f(a))\), the tangent line to the graph of \(f\) has slope \(f'(a)\). A line with that slope passing through that point has point-slope equation \(y-f(a)=f'(a)(x-a)\). Solving for \(y\) gives the linearization.
Thus, \(L(x)\) is not an unrelated approximation rule: it is the equation of the tangent line at the base point. At \(x=a\), it agrees exactly with the function, because \(L(a)=f(a)\). Its slope there also agrees with the function’s instantaneous rate of change. For inputs close to \(a\), the tangent line often tracks the graph well enough to give a useful estimate. The approximation sign matters: unless the function itself is linear, the estimate generally is not an exact equality.
A Reliable Procedure
Keep the base point fixed through the first calculations. In particular, do not evaluate the derivative at the target input just because that is the value you want to estimate.
Write down \(a\) and \(b\), so the signed change \(b-a\) is clear.
Evaluate \(f(a)\), showing the substitution and arithmetic.
Differentiate the function, then evaluate \(f'(a)\).
Use \(L(x)=f(a)+f'(a)(x-a)\), then substitute \(b\) and state the estimate.
When feasible, compare with the exact function value. Otherwise, check the arithmetic and whether the estimate is consistent with the function’s behavior.
Worked Examples from Function Formulas
Worked Example: Estimating a Polynomial Value
Use the linearization of \(f(x)=x^3-2x+1\) at \(a=2\) to estimate \(f(2.1)\), then compare with the exact value.
Step 1: Evaluate the function at the base point.
Step 2: Find the derivative and its value at the base point. Differentiating gives \(f'(x)=3x^2-2\), so
Step 3: Form the linearization.
Step 4: Substitute the target input. The change from the base point is \(2.1-2=0.1\), hence
Check: Direct evaluation gives \(f(2.1)=(2.1)^3-2(2.1)+1=9.261-4.2+1=6.061\). The tangent-line estimate is \(6\), compared with the exact value \(6.061\). The estimate is not exact, but the two values are close.
Worked Example: Estimating a Rational Function Value
For \(f(x)=\dfrac{x+1}{x+2}\), use the base point \(a=1\) to estimate \(f(1.04)\). The denominator is nonzero at \(x=1\), so the function is defined and differentiable there.
Step 1: Find the function value at \(a=1\).
Step 2: Differentiate and evaluate. The quotient rule gives
Therefore,
Step 3: Form the linearization and substitute the target.
Check: Direct evaluation gives \(f(1.04)=2.04/3.04=51/76\approx0.671053\). The difference between the estimate and the exact value is
The estimate is slightly above the exact value, and their difference is about \(0.0000585\). This check confirms both the arithmetic and that the estimate is close for this nearby input.
Worked Example: Estimating a Radical Value
Use \(f(x)=\sqrt{x}\) and the base point \(a=25\) to estimate \(\sqrt{25.16}\). The function is differentiable at \(25\), since \(25\) is positive.
Step 1: Evaluate at the base point.
Step 2: Differentiate and evaluate the slope. For \(x>0\), \(f'(x)=\dfrac{1}{2\sqrt{x}}\). Thus,
Step 3: Form and evaluate the linearization. The input is \(0.16\) above the base point, so
Check: Squaring the estimated value gives \(5.016^2=(5+0.016)^2=25+0.16+0.000256=25.160256\), slightly more than \(25.16\). Since both numbers are positive, this confirms that \(5.016\) is slightly greater than the true square root, as the estimate suggests. Direct calculation gives \(\sqrt{25.16}\approx5.015974\), close to \(5.016\).
A Common Mistake: Using the Target for the Slope
In the polynomial example, the base point is \(2\), while the target is \(2.1\). The slope in the linearization must be \(f'(2)\), not \(f'(2.1)\). A tempting but incorrect calculation is to keep \(f(2)=5\) and use the target-point derivative:
This does not use the tangent line at \(x=2\): its slope is \(f'(2)=10\). The correct calculation is \(5+10(0.1)=6\). The rule is simple: in \(f(a)+f'(a)(b-a)\), both function value and derivative are evaluated at the same base point \(a\). Only the change \(b-a\) uses the target input.
Interpreting and Reporting the Estimate
When the calculation is complete, report the result as an estimate, for example \(f(b)\approx L(b)\), rather than claiming an exact equality. It is useful to keep the exact form of the linearization through substitution and round only the final decimal. If the function is defined at the target and its exact value is easy to compute, comparing the two values provides a check; if not, checking the substitution, sign of \(b-a\), and arithmetic still helps catch errors.
The procedure applies across different formula types, but it does not make every estimate equally useful. The target should be nearby, and the function must have a derivative at the base point. If the function has a corner, vertical tangent, or is not defined at the chosen base point, this tangent-line method does not apply there. Also, being near the base point makes linearization a reasonable local model, not a promise that an estimate is exact.
Check Your Understanding
For each question, identify the base point before calculating the linear estimate.
- For \(f(x)=x^2+3x\), use \(a=2\) to estimate \(f(2.05)\). Write the linearization and show the substitution.
- For \(g(x)=\dfrac{1}{x}\), use \(a=2\) to estimate \(g(1.9)\). What is the sign of the change \(1.9-2\)?
- For \(h(x)=\sqrt{x+1}\), use \(a=3\) to estimate \(h(3.1)\). State the value and slope at the base point.
- In \(f(a)+f'(a)(b-a)\), which input determines the derivative, and which input determines the change?
- Why should a tangent-line estimate be written with an approximation sign rather than an equality in general?
