Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
$f(x)= x^2-5x+4$
$f'(x)=0=> 2x-5 =0$
$x= \large\frac{5}{2}$
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
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App