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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Prove that the function given by \(f (x) = x^3 – 3x^2 + 3x - 100\) is increasing in \(R.\)

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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 $f(x)=x^3-3x^2+3x-100$
Differentiating w.r.t $x$
$f'(x)=3x^2-6x+3$
$\qquad=3(x^2-2x+1)$
$\qquad=3(x-1)^2$
Step 2:
For any $x\in R,(x-1)^2 >0$
Thus $f'(x)$ is always positive in $R$.
Hence,the given function $f$ is increasing in $R$.
answered Jul 10, 2013 by sreemathi.v
 

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