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

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Question

(B) Given $f(x) = \frac{x - 1}{2x^2 - 7x + 5}$ for $x 
eq 1$.We can simplify the expression for $x 
eq 1$ by factoring the denominator:$2x^2 - 7x +

Vedclass · Accepted Answer

$-2/9$

Vedclass · Answer

(B) Given $f(x) = \frac{x - 1}{2x^2 - 7x + 5}$ for $x 
eq 1$.We can simplify the expression for $x 
eq 1$ by factoring the denominator:$2x^2 - 7x + 5 = 2x^2 - 2x - 5x + 5 = 2x(x - 1) - 5(x - 1) = (2x - 5)(x - 1)$.So,for $x 
eq 1$,$f(x) = \frac{x - 1}{(2x - 5)(x - 1)} = \frac{1}{2x - 5}$.By the definition of the derivative at $x = 1$:$f'(1) = \lim_{h 	o 0} \frac{f(1 + h) - f(1)}{h}$.Substituting $f(1 + h) = \frac{1}{2(1 + h) - 5} = \frac{1}{2h - 3}$ and $f(1) = -\frac{1}{3}$:$f'(1) = \lim_{h 	o 0} \frac{\frac{1}{2h - 3} - (-\frac{1}{3})}{h} = \lim_{h 	o 0} \frac{\frac{1}{2h - 3} + \frac{1}{3}}{h}$.$f'(1) = \lim_{h 	o 0} \frac{3 + (2h - 3)}{3h(2h - 3)} = \lim_{h 	o 0} \frac{2h}{3h(2h - 3)}$.$f'(1) = \lim_{h 	o 0} \frac{2}{3(2h - 3)} = \frac{2}{3(0 - 3)} = \frac{2}{-9} = -\frac{2}{9}$.

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'(1) = $

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