State when a function f(x) is said to be increasing

Solution:

Given:- Function $f(x)=f(x)=x^{2}-6 x+3$

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

(i) If $f^{\prime}(x)>0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$

Algorithm:-

(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

Here we have,

$f(x)=f(x)=x^{2}-6 x+3$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}-6 \mathrm{x}+3\right)$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-6$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2(\mathrm{x}-3)$

Here A function is said to be increasing on $[a, b]$ if $f(x)>0$

as given

$x \in[4,6]$

$\Rightarrow 4 \leq x \leq 6$

$\Rightarrow 1 \leq(x-3) \leq 3$

$\Rightarrow(x-3)>0$

$\Rightarrow 2(x-3)>0$

$\Rightarrow f^{\prime}(x)>0$

Hence, condition for $f(x)$ to be increasing

Thus $f(x)$ is increasing on interval $x \in[4,6]$