Question
Find the intervals in which the function f given by $f(x)=x^2-4 x+6$ is
a. increasing
b. decreasing
a. increasing
b. decreasing
✓
Answer
We have
$f(x)=x^2-4 x+6$
or $f ^{\prime}( x )=2 x -4$
Therefore, $f ^{\prime}( x )=0$ gives $x =2$. Now the point $x=2$ divides the real line into two disjoint intervals namely, $(-\infty, 2)$ and $(2, \infty)$
In the interval $(-\infty, 2), f ^{\prime}(x)=2 x-4<0$.
And in interval $(2, \infty), f^{\prime}(x)=2 x-4;0$
$\therefore$ (i) f is increasing in $(2, \infty)$
and (ii) $f$ is decreasing in $(-\infty, 2)$
$f(x)=x^2-4 x+6$
or $f ^{\prime}( x )=2 x -4$
Therefore, $f ^{\prime}( x )=0$ gives $x =2$. Now the point $x=2$ divides the real line into two disjoint intervals namely, $(-\infty, 2)$ and $(2, \infty)$
In the interval $(-\infty, 2), f ^{\prime}(x)=2 x-4<0$.
And in interval $(2, \infty), f^{\prime}(x)=2 x-4;0$
$\therefore$ (i) f is increasing in $(2, \infty)$
and (ii) $f$ is decreasing in $(-\infty, 2)$
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