Question

Use linear approximation to approximate √25.3 as follows.

Let f(x)=√x. The equation of the tangent line to f(x) at x=25 can be written in the form y=mx+b. Compute m,b.
Using this, find the approximation for √25.3.

Answer 1

The idea is to use the tangent line to
`f(x)=\sqrt x` at
`x=25` in order to approximate
`f(25.3)=√(25.3)`.

We have

`f(x)=\sqrt x\implies f(25)=√(25)=5`

`f'(x)=\frac1{2\sqrt x}\implies f'(25)=\frac1{10}`

so the linear approximation to
`f(x)` is

`L(x)=f(5)+f'(5)(x-5)=5+(x-5)/(10)=\frac x{10}+\frac92`

Hence
`m=\frac1{10}` and
`b=\frac92`.

Then

`f(25.3)\approx L(25.3)=(25.3)/(10)+\frac92=\boxed{7.03}`