Question

Use the denition of the derivative to find f 0 (3), where f (x) = 3x+5 / 2x−1

Answer 1

Answer:

$f'(3)= -\frac{13}{25}$

Step-by-step explanation:

We are asked to find $f'(3)$ of function $f(x)=\frac{3x+5}{2x-1}$ using definition of derivatives.

Limit definition of derivatives:

$f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Let us find $f(3+h)$ and $f(3)$.

$f(3+h)=\frac{3(3+h)+5}{2(3+h)-1}$

$f(3+h)=\frac{9+3h+5}{6+2h-1}\\\\f(3+h)=\frac{3h+14}{2h+5}$

$f(3)=\frac{3(3)+5}{2(3)-1}$

$f(3)=\frac{9+5}{6-1}\\\\f(3)=\frac{14}{5}$

Substituting these values in limit definition of derivatives, we will get:

$f'(3)= \lim_{h \to 0} \frac{f(3+h)-f(3)}{h}$

$f'(3)= \lim_{h \to 0} \frac{\frac{3h+14}{2h+5}-\frac{14}{5}}{h}$

Make a common denominator:

$f'(3)= \lim_{h \to 0} \frac{\frac{(3h+14)*5}{(2h+5)*5}-\frac{14(2h+5)}{5(2h+5)}}{h}$

$f'(3)= \lim_{h \to 0} \frac{\frac{5(3h+14)-14(2h+5)}{5(2h+5)}}{h}$

$f'(3)= \lim_{h \to 0} \frac{5(3h+14)-14(2h+5)}{5h(2h+5)}$

$f'(3)= \lim_{h \to 0} \frac{15h+70-28h-70}{5h(2h+5)}$

$f'(3)= \lim_{h \to 0} \frac{-13h}{5h(2h+5)}$

Cancel out h:

$f'(3)= \lim_{h \to 0} \frac{-13}{5(2h+5)}$

$f'(3)= \frac{-13}{5(2(0)+5)}$

$f'(3)= \frac{-13}{5(5)}$

$f'(3)= -\frac{13}{25}$

Therefore, $f'(3)= -\frac{13}{25}$.