Linear approximation

Near $x=a$, the curve $y=f(x)$ almost coincides with its tangent line at $(a,\ f(a))$. Linear approximation estimates values using that tangent line.

For example, with $f(x)=\sqrt{x}$, $a=4$ and $h=0.1$, we have $f'(4)=\frac{1}{4}$, so $\sqrt{4.1}\approx 2+\frac{1}{4}\cdot 0.1=2.025$ (the true value is $2.0248\ldots$). Written in $x$, it is $f(x)\approx f(a)+f'(a)(x-a)$, which is exactly the equation of the tangent line.

Examples

1. 01Approximate the change f(4+h) − f(4).$\displaystyle f(x)=x^{3}, \quad x=4$AnswerHide

   Answer$\displaystyle 48h$

   Steps

   1. The increase is slope × h
   2. $\displaystyle f'(x)$$\displaystyle = 3x^{2}$$\displaystyle f'(4)$$\displaystyle = 48$
   3. So
   4. $\displaystyle f(4+h)-f(4)$$\displaystyle \approx 48h$
2. 02Substitute h=0.1 into the linear approximation f(3)+f'(3)h.$\displaystyle f(3)=9, \quad f'(3)=6$AnswerHide

   Answer$\displaystyle 9.6$

   Steps

   1. f(3) is where you start; f'(3)×0.1 is the rise along the tangent
   2. $\displaystyle 9 + 6 \times 0.1$$\displaystyle = 9.6$
3. 03What is the linear approximation at this point?$\displaystyle f(x)=\sqrt{x}, \quad x=1$AnswerHide

   Answer$\displaystyle 1+\frac{1}{2}(x-1)$

   Steps

   1. The value at the point
   2. $\displaystyle \sqrt{1}$$\displaystyle = 1$
   3. The slope there
   4. $\displaystyle \frac{1}{2\sqrt{1}}$$\displaystyle = \frac{1}{2}$
   5. Put both into the tangent-line form
   6. $\displaystyle y$$\displaystyle = 1+\frac{1}{2}(x-1)$
4. 04What goes in □?$\displaystyle f(a+h) \approx f(a) + \square$AnswerHide

   Answer$\displaystyle f'(a)h$

   Steps

   1. Along the tangent, the rise is slope × run
   2. $\displaystyle f(a+h)$$\displaystyle \approx f(a) + f'(a)h$
   3. Without h, any run would give the same rise
5. 05Substitute h=0.1 into the linear approximation f(5)+f'(5)h.$\displaystyle f(5)=25, \quad f'(5)=10$AnswerHide

   Answer$\displaystyle 26.0$

   Steps

   1. f(5) is where you start; f'(5)×0.1 is the rise along the tangent
   2. $\displaystyle 25 + 10 \times 0.1$$\displaystyle = 26.0$
6. 06Substitute h=0.1 into the linear approximation f(4)+f'(4)h.$\displaystyle f(4)=16, \quad f'(4)=8$AnswerHide

   Answer$\displaystyle 16.8$

   Steps

   1. f(4) is where you start; f'(4)×0.1 is the rise along the tangent
   2. $\displaystyle 16 + 8 \times 0.1$$\displaystyle = 16.8$

FAQ

Q1How accurate is it?AnswerHide

A

The smaller $h$ is, the smaller the error; making $h$ larger makes the error larger. For $f(x)=x^2$, $f(a+h)-\{f(a)+f'(a)h\}=h^2$, so the error is $h^2$.

Q2How do I approximate just the change $f(a+h)-f(a)$?AnswerHide

A

Move $f(a)$ across: $f(a+h)-f(a)\approx f'(a)h$. For $f(x)=x^3$ and $a=4$, $f'(4)=48$, so the change is about $48h$.