given $f(x)=3sqrt{x - 1}$, find $f(x)$ using the limit definition of the derivative.\n$f(x)=\nquestion help…

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative is $f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=3\sqrt{x - 1}$, then $f(x + h)=3\sqrt{(x + h)-1}=3\sqrt{x+h - 1}$.

Step2: Substitute into the formula

$\frac{f(x + h)-f(x)}{h}=\frac{3\sqrt{x + h-1}-3\sqrt{x - 1}}{h}$.

Step3: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator $3\sqrt{x + h-1}+3\sqrt{x - 1}$. [ \begin{align*} \frac{3\sqrt{x + h-1}-3\sqrt{x - 1}}{h}\times\frac{3\sqrt{x + h-1}+3\sqrt{x - 1}}{3\sqrt{x + h-1}+3\sqrt{x - 1}}&=\frac{9(x + h-1)-9(x - 1)}{h(3\sqrt{x + h-1}+3\sqrt{x - 1})}\ &=\frac{9x+9h - 9-9x + 9}{h(3\sqrt{x + h-1}+3\sqrt{x - 1})}\ &=\frac{9h}{h(3\sqrt{x + h-1}+3\sqrt{x - 1})} \end{align*} ]

Step4: Simplify the expression

Cancel out the $h$ terms: $\frac{9h}{h(3\sqrt{x + h-1}+3\sqrt{x - 1})}=\frac{9}{3\sqrt{x + h-1}+3\sqrt{x - 1}}$.

Step5: Take the limit as $h\rightarrow0$

$f^{\prime}(x)=\lim_{h\rightarrow0}\frac{9}{3\sqrt{x + h-1}+3\sqrt{x - 1}}=\frac{9}{3\sqrt{x-1}+3\sqrt{x - 1}}=\frac{9}{6\sqrt{x - 1}}=\frac{3}{2\sqrt{x - 1}}$.

Answer:

$\frac{3}{2\sqrt{x - 1}}$