logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

Prove that the function \(f\) given by \(f (x) = x^2 – x + 1\) is neither strictly increasing nor strictly decreasing on \((– 1, 1).\)

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
  • If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
  • A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
  • The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
  • The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Let $f(x)=x^2-x+1$
Differentiating w.r.t $x$ we get,
$f'(x)=2x-1$
$\quad=2(x-\large\frac{1}{2})$
Clearly $-1 < x < \large\frac{1}{2}$
$\Rightarrow f'(x) <0$
Step 2:
$\large\frac{1}{2}$$< x < 1\Rightarrow x-\large\frac{1}{2}$$ > 0$
$\Rightarrow 2(x-\large\frac{1}{2})$$ > 0\Rightarrow f'(x) > 0$
Thus $f'(x)$ does not have the same sign through out the interval $(-1,1)$
Hence it is neither increasing nor decreasing on $(-1,1)$
answered Jul 9, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...