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Answer:

$i.\: x=3 \\ ii.\: co-ordinate (\frac{5}{2},\frac{1}{4})$

Given:

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

To find:

We have to find the value of x, also we have to find co-ordinate of the point on the given curve where the normal is parallel to the line y = x+5.

Hint:

1. First we will find critical point then find f’(x) for all the value of x.
2. Find m1 and m2 from the curve then find points

Solution:

We have

\begin{aligned} &f^{\prime}(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) \end{aligned}

For f(x) let us find critical point, we must have

\begin{aligned} &\Rightarrow f^{\prime}(x)=0 \\ &\Rightarrow 2(x-3)=0 \\ &\Rightarrow x-3=0 \\ &\Rightarrow x=3 \\ &\text { Clearly, } f^{\prime}(x)>0 i f x>3 \text { and } f^{\prime}(x)<0 \text { if } x<3 \end{aligned}

Thus f(x) increases on 3, $\infty$ and f(x) is decreasing on interval x ∈ (-$\infty$, 3)

Now,let us find co-ordinates of point on the equation of curve is

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

Slope of this curve is given by

\begin{aligned} &\Rightarrow m_{1}=\frac{d y}{d x} \\ &\Rightarrow m_{1}=\frac{d}{d x}\left(x^{2}-6 x+9\right) \\ &\Rightarrow m_{1}=2x-6 \end{aligned}

Equation of line is y=x+5

Slope of this curve is

\begin{aligned} &\Rightarrow m_{2}=\frac{d y}{d x} \\ &\Rightarrow m_{2}=\frac{d}{d x}(x+5) \\ &\Rightarrow m_{2}=1 \end{aligned}

Since slope of curve is parallel to line.

Therefore, they follow the relation

\begin{aligned} &\Rightarrow-\frac{1}{m_{1}}=m_{2} \\ &\Rightarrow-\frac{1}{2 x-6}=1 \\ &\Rightarrow 2 x-6=-1 \\ &\Rightarrow x=\frac{5}{2} \end{aligned}

Thus putting the value of x in equation of curve

We get

\begin{aligned} &\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} \end{aligned}

Hence the required co-ordinate is

$(\frac{5}{2},\frac{1}{4})$

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