Question

[SCREENSHOT INCLUDED] If f(x) = sqrt(2x+1) then f '(4) =

Answer 1

Using the definition,

$f'(4) = \displaystyle \lim_{x\to4} \frac{f(x)-f(4)}{x-4}$

The numerator in the difference quotient is

$f(x) - f(4) = \sqrt{2x+1} - \sqrt{2\cdot4+1} = \sqrt{2x+1}-3$

Multiply the numerator and denominator by the conjugate of this expression,

$\displaystyle \frac{\sqrt{2x+1}-3}{x-4} \cdot \dfrac{\sqrt{2x+1}+3}{\sqrt{2x+1}+3}$

The modified numerator reduces to a difference of squares,

$\displaystyle \frac{\left(\sqrt{2x+1}\right)^2-3^2}{(x-4)\left(\sqrt{2x+1}+3\right)}$

Simplify this to get

$\displaystyle \frac{2x+1-9}{(x-4)\left(\sqrt{2x+1}+3\right)} = \frac{2(x-4)}{(x-4)\left(\sqrt{2x+1}+3\right)}$

Since x is approaching 4, it never actually takes on the value of 4, so (x - 4)/(x - 4) reduces to 1. Then the limit is equivalent to

$f'(4) = \displaystyle \lim_{x\to4} \frac{f(x)-f(4)}{x-4} = \lim_{x\to4}\frac2{\sqrt{2x+1}+3}$

and the remaining limand is continuous at x = 4, so that

$f'(4) = \dfrac2{\sqrt{2\cdot4+1}+3} = \dfrac26 = \boxed{\dfrac13}$