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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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) = $

Similar Questions

If $f(x) = \begin{cases} \frac{x^2 \ln \cos x}{\ln(1 + x^2)}, & x \neq 0 \\ 0, & x = 0 \end{cases}$,then $f(x)$ is

Let $S$ be the set of points where the function $f(x) = |2 - |x - 3||, x \in R,$ is not differentiable. Then $\sum_{x \in S} f(f(x))$ is equal to

Let $f(x) = |x|$. Then which of the following is true?

If $f(x) = \begin{cases} x \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$,then the correct statement is:

Let $f(x) = \begin{cases} x + 1, & \text{when } x < 2 \\ 2x - 1, & \text{when } x \ge 2 \end{cases}$,then $f'(2) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module