Question

question 12
suppose that $f$ is a function given as $f(x) = \frac{1}{2x - 5}$.
simplify the expression $f(x+h)$.
$f(x+h) = \square$
simplify the difference quotient, $\frac{f(x+h)-f(x)}{h}$.
$\frac{f(x+h)-f(x)}{h} = \square$
the derivative of the function at $x$ is the limit of the difference quotient as $h$ approaches zero.
$f(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \square$

Explanation:

Step1: Substitute $x+h$ into $f(x)$

$f(x+h)=\frac{1}{2(x+h)-5}=\frac{1}{2x+2h-5}$

Step2: Write difference quotient expression

$\frac{f(x+h)-f(x)}{h}=\frac{\frac{1}{2x+2h-5}-\frac{1}{2x-5}}{h}$

Step3: Combine fractions in numerator

$\frac{\frac{(2x-5)-(2x+2h-5)}{(2x+2h-5)(2x-5)}}{h}=\frac{\frac{-2h}{(2x+2h-5)(2x-5)}}{h}$

Step4: Cancel $h$ from numerator/denominator

$\frac{-2}{(2x+2h-5)(2x-5)}$

Step5: Take limit as $h\to0$

$\lim_{h\to0}\frac{-2}{(2x+2h-5)(2x-5)}=\frac{-2}{(2x-5)^2}$

Answer:

$f(x+h)=\boldsymbol{\frac{1}{2x+2h-5}}$
$\frac{f(x+h)-f(x)}{h}=\boldsymbol{\frac{-2}{(2x+2h-5)(2x-5)}}$
$f'(x)=\boldsymbol{\frac{-2}{(2x-5)^2}}$