linear approximation of a function at a number

Question

question: Find linear approximation of the function f(x) = √x at x=9. Use it to approximate √9.1

linearisation formula L(x) = f(a) + f'(a)(x-a)

In this question, a=9 right? and when we substitute everything we should get f(a)= √9 = 3 f'(a)=1/6 x-a=x-9

Further substituting, for estimating √9.1 we just have to put 9.1 in place of x right?? Or have I understood this completely wrong? L(x)=3 + 1/3(x-9) = 3 + 1/6(9.1-9) =3.0166667

I realise that this is probably a dumb question but can someone please confirm?

Answer 1

The function $$f(x)=\sqrt x$$ has a derivative $$f'(x)=\frac1{2\sqrt x}.$$ Using these two equations, we know that at $x=9$, the function has a value of $f(9)=3$ and a slope of $f'(9)=\dfrac16$.

Linear approximation tells us that in the neighbourhood of $x=9$, the values of the function can be approximated by the tangent line at $x=9$. Since the slope of the tangent line is $\dfrac16$, we have $$f(9+\Delta x)\approx f(9)+f'(9)\Delta x=3+\frac16\times\frac1{10}=\frac{181}{60}\doteq3.01\bar6$$ as you have written.