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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the intervals in which the following functions are strictly increasing or decreasing: $(b) \;10-6x-2x^2$

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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:
Given :$f(x)=10-6x-2x^2$
Differentiating w.r.t $x$ we get,
$f'(x)=-6-4x$
$\qquad=-2(3+2x)$
Step 2:
$f(x)$ is increasing,if $f'(x)>0$
$\Rightarrow -6-4x>0$
$\Rightarrow -4x>6$
$\Rightarrow x<\large\frac{-3}{2}$
Step 3:
$f(x)$ is decreasing if $f'(x) < 0$
(i.e) if $-6-4x<0$
$\Rightarrow x<\large\frac{-3}{2}$
Hence $f(x)$ is increasing for $x<\large\frac{-3}{2}$ (i.e)in the interval $(-\infty,-\large\frac{3}{2})$ and decreasing for $x>\large\frac{-3}{2}$ (i.e) $(\large\frac{-3}{2}$$,\infty)$.
answered Jul 9, 2013 by sreemathi.v
 

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