Why Linearize a Power Function?
Some powers are easy to evaluate exactly, while nearby ones may not be. For instance, \(4^3\) is immediate, but evaluating \(4.02^3\) by repeated multiplication takes more work. A linearization uses the value and slope of a function at a nearby, convenient point to produce a simpler estimate.
For a power function, the method applies to both integer and fractional exponents, provided the input and the base point are positive. The estimate is most useful when the input is close to the base point. It is an approximation, not usually an exact replacement for the original power.
The Linearization of a Power Function
Let \(p\) be an integer or rational number, and consider \(f(x)=x^p\) for positive \(x\). At a positive base point \(a\), the linearization is the tangent-line model built from \(f(a)\) and \(f'(a)\). The power rule gives \(f'(x)=p x^{p-1}\), so the linearization has a direct formula.
The difference \(x-a\) records how far the input is from the base point. The coefficient \(p a^{p-1}\) is the slope of the power function at that point. To estimate \(x^p\), substitute the desired input for \(x\) in \(L_a(x)\).
Why the Formula Works
The tangent-line model for a differentiable function \(f\) at \(a\) is \(f(a)+f'(a)(x-a)\). It matches the function's value at \(a\) and has the same slope there. For the power function, \(f(a)=a^p\) and \(f'(a)=p a^{p-1}\). Substituting these two quantities into the tangent-line model gives the stated formula.
In particular, setting \(x=a\) makes the change \(x-a\) equal to zero, so \(L_a(a)=a^p\). The slope of \(L_a\) is the constant \(p a^{p-1}\), which is also \(f'(a)\). Thus the line agrees with the power function in both value and slope at the base point. These two matches are the reason it is a useful local model.
Worked Example: Estimating a Nearby Cube
Worked Example: Estimate \(1.02^3\)
Use the nearby convenient base point \(a=1\). Here \(p=3\), so \(f(x)=x^3\) and \(f'(x)=3x^2\).
First evaluate the function and its derivative at the base point:
Therefore the linearization is
Substitute \(x=1.02\). The input change is \(1.02-1=0.02\), so
Thus \(1.02^3\) is estimated as \(1.06\). To check the construction, \(L_1(1)=1\), matching \(1^3\), and the slope of \(L_1\) is \(3\), matching \(f'(1)=3\). Direct multiplication gives \(1.02^2=1.0404\), then \(1.0404(1.02)=1.061208\). The estimate is close but not exact; its difference from the directly computed value is \(1.06-1.061208=-0.001208\).
Integer Exponents at Other Base Points
The base point need not be \(1\). A useful choice may be a nearby number whose power and derivative are easy to evaluate. The same formula works for any positive base point, including for negative integer exponents, as long as the inputs considered are positive.
Worked Example: Estimate \(3.98^2\)
Take \(f(x)=x^2\), \(p=2\), and \(a=4\). The function value and derivative at the base point are
Thus
For \(x=3.98\), the change from the base point is \(3.98-4=-0.02\). Substitution gives
So the estimate is \(3.98^2\approx15.84\). Check the model at the base point: \(L_4(4)=16\), equal to \(4^2\), and its slope is \(8\), equal to \(f'(4)\). Direct multiplication gives \(3.98^2=15.8404\), so the estimate differs from the exact value by \(15.84-15.8404=-0.0004\).
Fractional Exponents
The formula also applies when \(p\) is a fraction. A positive input is important here: for a general rational exponent, restricting to \(x>0\) gives a consistent real-valued power function and lets us use the power rule \(f'(x)=p x^{p-1}\). The derivative calculation and the linearization steps are otherwise the same as for integer exponents.
Worked Example: Estimate \(4.04^{3/2}\)
Let \(f(x)=x^{3/2}\) and choose \(a=4\). The base point is convenient because \(4^{1/2}=2\). First compute the function value:
By the power rule,
At the base point,
Therefore
At \(x=4.04\), the input change is \(4.04-4=0.04\). Hence
The linearization estimates \(4.04^{3/2}\) as \(8.12\). To check the model itself, \(L_4(4)=8=4^{3/2}\), and its slope is \(3=f'(4)\). The calculation uses the fractional-exponent power rule at the positive input \(4\); it does not claim that \(8.12\) is the exact value of the power.
A Reliable Calculation Process
Writing the formula before substituting values helps keep the base point, exponent, and input change distinct. In particular, the coefficient of \(x-a\) is the derivative evaluated at \(a\), not the exponent alone.
Write \(f(x)=x^p\), identify \(p\), and choose a positive base point \(a\) near the input of interest.
Compute \(f(a)=a^p\) and \(f'(a)=p a^{p-1}\), keeping the exponent arithmetic explicit.
Use \(L_a(x)=a^p+p a^{p-1}(x-a)\).
Calculate the change \(x-a\) and evaluate the resulting linear expression.
Confirm that the model returns \(a^p\) at \(x=a\), and that its slope is \(p a^{p-1}\).
Common Mistake: Using the Wrong Slope
A frequent error is to use \(p a^p\) as the coefficient of \(x-a\). The derivative of \(x^p\) is \(p x^{p-1}\), so evaluating it at \(a\) gives \(p a^{p-1}\), with exponent \(p-1\), not \(p\).
Worked Example: Correcting a Slope Error
Suppose we want to estimate \(2.1^3\) using \(a=2\). A wrong calculation might use the slope \(3(2^3)=24\), giving
This is incorrect because \(24\) is not the derivative at \(2\). The correct derivative is \(f'(x)=3x^2\), so
The correct linearization is \(L_2(x)=8+12(x-2)\). Therefore
Check: \(L_2(2)=8=2^3\), and the line's slope is \(12=f'(2)\). Direct multiplication gives \(2.1^2=4.41\) and \(4.41(2.1)=9.261\); the linear estimate is not exact, but it comes from the correct tangent slope.
When to Use the Estimate
A linearization is a local approximation: it uses information at one point to estimate values nearby. As the input moves closer to \(a\), the tangent-line model captures the function's first-order change more directly. If the requested input is far from the base point, the estimate may be poor, and the formula alone does not provide a numerical error bound.
For a good calculation, check three things: the base point is positive and near the target input; the derivative is evaluated at that base point; and the change is entered as \(x-a\), including its sign. A negative change means the target lies below the base point, as in the estimate of \(3.98^2\) near \(4^2\).
Check Your Understanding
For each question, identify the base point, use the derivative \(p x^{p-1}\), and keep the input change explicit.
- Write the linearization of \(x^4\) at \(a=1\), then use it to estimate \(1.03^4\).
- Find the linearization of \(x^{1/2}\) at \(a=9\). What estimate does it give for \(9.1^{1/2}\)?
- For \(f(x)=x^{2/3}\) at \(a=8\), calculate \(f(a)\), \(f'(a)\), and the resulting linearization.
- Explain why \(p a^p\) is not the correct slope for linearizing \(x^p\) at \(a\).
- What two checks confirm that a proposed linearization has the correct value and slope at its base point?
