Respuesta :
Answer:
[tex]f'(3)= -\frac{13}{25}[/tex]
Step-by-step explanation:
We are asked to find [tex]f'(3)[/tex] of function [tex]f(x)=\frac{3x+5}{2x-1}[/tex] using definition of derivatives.
Limit definition of derivatives:
[tex]f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]
Let us find [tex]f(3+h)[/tex] and [tex]f(3)[/tex].
[tex]f(3+h)=\frac{3(3+h)+5}{2(3+h)-1}[/tex]
[tex]f(3+h)=\frac{9+3h+5}{6+2h-1}\\\\f(3+h)=\frac{3h+14}{2h+5}[/tex]
[tex]f(3)=\frac{3(3)+5}{2(3)-1}[/tex]
[tex]f(3)=\frac{9+5}{6-1}\\\\f(3)=\frac{14}{5}[/tex]
Substituting these values in limit definition of derivatives, we will get:
[tex]f'(3)= \lim_{h \to 0} \frac{f(3+h)-f(3)}{h}[/tex]
[tex]f'(3)= \lim_{h \to 0} \frac{\frac{3h+14}{2h+5}-\frac{14}{5}}{h}[/tex]
Make a common denominator:
[tex]f'(3)= \lim_{h \to 0} \frac{\frac{(3h+14)*5}{(2h+5)*5}-\frac{14(2h+5)}{5(2h+5)}}{h}[/tex]
[tex]f'(3)= \lim_{h \to 0} \frac{\frac{5(3h+14)-14(2h+5)}{5(2h+5)}}{h}[/tex]
[tex]f'(3)= \lim_{h \to 0} \frac{5(3h+14)-14(2h+5)}{5h(2h+5)}[/tex]
[tex]f'(3)= \lim_{h \to 0} \frac{15h+70-28h-70}{5h(2h+5)}[/tex]
[tex]f'(3)= \lim_{h \to 0} \frac{-13h}{5h(2h+5)}[/tex]
Cancel out h:
[tex]f'(3)= \lim_{h \to 0} \frac{-13}{5(2h+5)}[/tex]
[tex]f'(3)= \frac{-13}{5(2(0)+5)}[/tex]
[tex]f'(3)= \frac{-13}{5(5)}[/tex]
[tex]f'(3)= -\frac{13}{25}[/tex]
Therefore, [tex]f'(3)= -\frac{13}{25}[/tex].
