Determine the values of x for which the function

Determine the values of $x$ for which the function $f(x)=x^{2}-6 x+9$ is increasing or decreasing. Also, find the coordinates of the point on the curve $y=x^{2}-6 x+9$ where the normal is parallel to the liney $=x+5$.


Given:- Function $f(x)=x^{2}-6 x+9$ and a line parallel to $y=x+5$

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 $(a, b)$


(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,

$f(x)=x^{2}-6 x+9$

$\Rightarrow f(x)=\frac{d}{d x}\left(x^{2}-6 x+9\right)$

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

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

For $f(x)$ lets find critical point, we must have

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

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


$\Rightarrow x=3$

clearly, $f^{\prime}(x)>0$ if $x>3$

and $f^{\prime}(x)<0$ if $x<3$

Thus, $f(x)$ increases on $(3, \infty)$

and $f(x)$ is decreasing on interval $x \in(-\infty, 3)$

Now, lets find coordinates of point

Equation of curve is

$f(x)=x^{2}-6 x+9$

slope of this curve is given by

$\Rightarrow \mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow \mathrm{m}_{1}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}-6 \mathrm{x}+9\right)$

$\Rightarrow \mathrm{m}_{1}=2 \mathrm{x}-6$

and Equation of line is


slope of this curve is given by

$\Rightarrow \mathrm{m}_{2}=\frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow \mathrm{m}_{2}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+5)$

$\Rightarrow \mathrm{m}_{2}=1$

Since slope of curve (i.e slope of its normal) is parallel to line

Therefore, they follow the relation

$\Rightarrow \frac{-1}{m_{1}}=m_{2}$

$\Rightarrow \frac{-1}{2 x-6}=1$

$\Rightarrow 2 x-6=-1$

$\Rightarrow x=\frac{5}{2}$

Thus putting the value of $\mathrm{x}$ in equation of curve, we get

$\Rightarrow y=x^{2}-6 x+9$

$\Rightarrow y=\left(\frac{5}{2}\right)^{2}-6\left(\frac{5}{2}\right)+9$

$\Rightarrow y=\frac{25}{4}-15+9$

$\Rightarrow y=\frac{25}{4}-6$

$\Rightarrow y=\frac{1}{4}$

Thus the required coordinates is $\left(\frac{5}{2}, \frac{1}{4}\right)$


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