If f(x) = begin{cases} frac{x-1}{2x^2-7x+5}, | vedclass

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(B) Given $f(x) = \frac{x-1}{2x^2-7x+5}$ for $x 
eq 1$.Factoring the denominator: $2x^2-7x+5 = 2x^2-2x-5x+5 = 2x(x-1)-5(x-1) = (2x-5)(x-1)$.Thus,for

Vedclass · Accepted Answer

$-\frac{2}{9}$

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(B) Given $f(x) = \frac{x-1}{2x^2-7x+5}$ for $x 
eq 1$.Factoring the denominator: $2x^2-7x+5 = 2x^2-2x-5x+5 = 2x(x-1)-5(x-1) = (2x-5)(x-1)$.Thus,for $x 
eq 1$,$f(x) = \frac{x-1}{(2x-5)(x-1)} = \frac{1}{2x-5}$.The function is $f(x) = \begin{cases} \frac{1}{2x-5}, & x 
eq 1 \ -\frac{1}{3}, & x=1 \end{cases}$.By the definition of the derivative,$f^{\prime}(1) = \lim_{h 	o 0} \frac{f(1+h)-f(1)}{h}$.$f^{\prime}(1) = \lim_{h 	o 0} \frac{\frac{1}{2(1+h)-5} - (-\frac{1}{3})}{h} = \lim_{h 	o 0} \frac{\frac{1}{2h-3} + \frac{1}{3}}{h}$.$f^{\prime}(1) = \lim_{h 	o 0} \frac{3 + (2h-3)}{3h(2h-3)} = \lim_{h 	o 0} \frac{2h}{3h(2h-3)}$.$f^{\prime}(1) = \lim_{h 	o 0} \frac{2}{3(2h-3)} = \frac{2}{3(-3)} = -\frac{2}{9}$.

If $f(x) = \begin{cases} \frac{x-1}{2x^2-7x+5}, & \text{for } x \neq 1 \\ -\frac{1}{3}, & \text{for } x=1 \end{cases}$,then $f^{\prime}(1)$ is equal to:

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