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Find the intervals in which the function $f$ given by $f (x) = 2x^2 – 3x$ is $ (b)\; strictly\; decreasing$

This is (b)part of multipart q4

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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 :$f(x)=2x^2-3x$
Differentiating w.r.t $x$ we get,
$f'(x)=4x-3$
$f'(x)=0$
$\Rightarrow 4x-3=0$
$x=\large\frac{3}{4}$
Step 2:
This point $x=\large\frac{3}{4}$ divides the real number line into two intervals (i.e) $(\infty,\large\frac{3}{4})$ and $(\large\frac{3}{4},$$\infty)$
In the interval $(-\infty,\large\frac{3}{4})$ $f'(x)$ is negative .Hence it is strictly decreasing in $(-\infty,\large\frac{3}{4})$
answered Aug 16, 2013 by sreemathi.v
 

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