Using linear approximation to approximate $\sqrt{81.3}$

Question

Use linear approximaiton to approximate $\sqrt{81.3}$ as follows: Let $f(x)=\sqrt{x}$. The equation of the tangent line to $f(x)$ at $x=81$ can be written in the form $y=mx+b$ where $m$ is:____ and where $b$ is:_____

I calculated:

$f'(x)=1/2x^{-1/2}= 1/18=m$

$f(x)=\sqrt{81}=9=b$, why is this wrong for $b$? Am I missing something?

Answer 1

Recall point-slope form at a point $(x_0,y_0)$: $$y = m(x-x_0) + y_0.$$ Now, recasting in this in Calculus terms, we have the line tangent to $f$, here $\ell(x)$, at the point $(x_0,f(x_0))$: $$\ell(x) = f'(x_0)(x-x_0) + f(x_0).$$ You correctly found $f'(x_0)$ and $f(x_0)$; but notice that this information defines the line in point-slope form, i.e., $$\ell(x) = \frac{1}{18}(x - 81) + 9.$$ Do you see how this is not yet in slope-intercept form? ($y=mx + b$). Some simplification is in order.

Answer 2

Point : $(81, 9)$
Slope : $\dfrac{1}{18}$

Point-slope form of line would be : $$y-9 = \dfrac{1}{18}(x-81)$$

Change it into slope-intercept form