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The function $f(x)=x^{2}-5x+4$ is increasing in

\[\begin{array}{1 1}(1)(-\infty , 1)&(2)(1 , 4 )\\(3)(4 , \infty )&(4)every where\end{array}\]

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$f(x)= x^2-5x+4$
$f'(x)=0=> 2x-5 =0$
$x= \large\frac{5}{2}$
When $x < \large\frac{5}{2}$ say $x=2$
$f'(x) =4-5 <0$
$\therefore f'(x) $ is -ve i $(-\infty, \large\frac{5}{2})$
When $x > \large\frac{5}{2}$ say $x=3$
$f'(x) =6-5 >0$
$f'(x)$ is +ve i $(3 , \infty)$
$f(x)$ is a increasing function i $(4, \infty)$
Hence 3 is the correct answer
answered May 20, 2014 by meena.p
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