Why Use a Tangent-Line Model?
A function can be complicated to evaluate at a particular input. Near a point where the function is differentiable, however, its graph has a well-defined tangent line. That line gives a simpler way to estimate nearby function values.
The estimate is local: it uses information at one base point and is intended for inputs near that point. It is not generally an exact replacement for the function. The closer the input is to the base point, the more reasonable it is to expect the tangent line to reflect the graph’s behavior there.
Definition of the Tangent-Line Model
Let \(f\) be differentiable at \(a\), and suppose \(f\) is defined in a neighborhood of \(a\). The number \(a\) is the base point. The tangent line to the graph of \(f\) at \((a,f(a))\) has slope \(f'(a)\). Its equation defines the tangent-line model:
The model uses three pieces of information: the base point \(a\), the function value \(f(a)\), and the derivative \(f'(a)\). For an input \(x\) near \(a\), \(L(x)\) is used as an estimate of \(f(x)\).
The formula is the point-slope equation of the line through \((a,f(a))\) with slope \(f'(a)\). In particular, the model agrees with the function at the base point:
Its slope is also the derivative at the base point, since \(L'(x)=f'(a)\), a constant. Thus the line matches both the function’s value and its slope at \(a\). The model is a line in \(x\), even though its standard name is a linearization.
How to Use the Model
To estimate \(f(x)\) near \(a\), first find \(f(a)\) and \(f'(a)\). Substitute them into the model, then evaluate the model at the desired input. It is useful to keep the difference \(x-a\) visible: it shows how far, and in which direction, the new input lies from the base point.
Identify the point \(a\) where the function value and derivative are known or easy to compute.
Calculate \(f(a)\) and \(f'(a)\).
Use \(L(x)=f(a)+f'(a)(x-a)\).
Use \(L(x)\) as the estimate for \(f(x)\), and label it as an approximation unless equality is known.
Worked Examples
Worked Example: Estimating a Value Just to the Right of the Base Point
Let \(f(x)=x^3-2x+1\). Use the tangent-line model at \(a=1\) to estimate \(f(1.04)\), then compare with the exact value.
First evaluate the function at the base point:
Differentiate and evaluate the derivative at \(1\):
Substitute \(a=1\), \(f(1)=0\), and \(f'(1)=1\) into the model:
At \(x=1.04\), the model gives:
Therefore, \(f(1.04)\approx 0.04\). To compare, compute the exact value:
The exact value is \(0.044864\), while the tangent-line estimate is \(0.04\). The estimate is close but not exact. Checking the model directly confirms the calculation: \(L(1.04)=1.04-1=0.04\).
Worked Example: Estimating a Value to the Left of the Base Point
Let \(g(x)=x^3-3x+4\). Use the tangent-line model at \(a=2\) to estimate \(g(1.98)\), and check the estimate against the exact function value.
Evaluate the function at \(2\):
The derivative and its value at \(2\) are:
The tangent-line model is:
Since \(1.98-2=-0.02\), the model gives:
Thus \(g(1.98)\approx 5.82\). For comparison, compute the exact value:
The exact value is \(5.822392\), close to the estimate \(5.82\). Substitution into the model checks the estimate: \(6+9(1.98-2)=6-0.18=5.82\). The negative input change correctly contributes a negative amount because the slope is positive.
Worked Example: Using a Known Value and Derivative
Suppose \(h\) is differentiable at \(a=4\), with \(h(4)=10\) and \(h'(4)=-\frac{3}{2}\). Use the model to estimate \(h(4.06)\).
The tangent-line model at \(4\) is:
The input is \(0.06\) greater than the base point. Substitute \(x=4.06\):
Therefore, the model estimates \(h(4.06)\) as \(9.91\). Checking by substituting into the model again gives \(10-\frac{3}{2}(0.06)=9.91\). Since no formula or exact value for \(h(4.06)\) is given, this calculation checks the model evaluation, not the accuracy of the estimate against the unknown function value.
Why the Tangent Line Is a Reasonable Local Model
Differentiability at \(a\) means that the derivative exists as a finite limit:
This limit describes the function’s rate of change near \(a\). The tangent line passes through \((a,f(a))\) and has slope \(f'(a)\), so its change from \(x=a\) to a nearby input \(x\) is \(f'(a)(x-a)\). Adding that predicted change to the starting value \(f(a)\) gives \(f(a)+f'(a)(x-a)\), exactly the model \(L(x)\).
This reasoning explains why the model captures the function’s local value and slope. Differentiability is essential: if the derivative does not exist at the base point, this construction does not provide a tangent-line model with a finite slope there. Even when the derivative exists, the model is an estimate at other inputs, not a claim that the function itself is a line.
Reading the Model’s Direction and Range
The sign of \(f'(a)\) describes the direction of the model’s change as \(x\) moves from \(a\). If \(x-a\) is positive, the model’s change from \(f(a)\) has the sign of \(f'(a)\). If \(x-a\) is negative, the sign of that change is reversed. If \(f'(a)=0\), the tangent-line model is the constant function \(L(x)=f(a)\).
These statements describe the model’s change, not necessarily the exact change in \(f(x)\) at every nearby input. The function can curve away from its tangent line. The model is most naturally used for inputs near the base point; moving far from \(a\) gives less reason to expect the local line to remain a useful estimate.
A Common Mistake: Using the Wrong Input Difference
A frequent error is to replace \(x-a\) with \(x\). The model depends on the change from the base point, not on the input alone. For example, in the second worked example the base point is \(2\), the slope is \(9\), and the target input is \(1.98\). The correct calculation uses \(1.98-2=-0.02\).
This wrong expression treats the input itself as the change from the base point. The correction is to subtract the base point before multiplying by the slope:
The base-point check is a useful safeguard. A correct model must give \(L(a)=f(a)\). Here \(L(2)=6+9(2-2)=6\). An expression using \(6+9x\) would fail that check, since its value at \(2\) is \(24\), not \(6\).
Summary
For a function differentiable at \(a\), its tangent-line model is \(L(x)=f(a)+f'(a)(x-a)\). The point \(a\) supplies the starting value and the slope; the difference \(x-a\) supplies the input change. Evaluate \(L(x)\) to estimate \(f(x)\) near \(a\), and remember that the estimate is generally not an exact function value.
Check Your Understanding
For each question, identify which information comes from the base point and which comes from the input being estimated.
- If \(f(3)=7\) and \(f'(3)=2\), write the tangent-line model at \(a=3\).
- Using that model, estimate \(f(3.05)\). Show the input difference \(x-a\).
- If \(g(5)=4\) and \(g'(5)=-3\), what does the model predict for \(g(4.9)\)?
- Why must the model agree with the function at the base point?
- Does differentiability at \(a\) imply that the tangent-line model equals the function at every input near \(a\)? Explain.
