Question

Determine the values of x for which the function f(x) = x² − 6x + 9 is increasing or decreasing. Also, find the coordinates of the point on the curve y = x² − 6x + 9 where the normal is parallel to the line y = x + 5.

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ f\left(x\right)=x^2-6x+9 \\ f\text{'}\left(x\right)=2x-6 \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 2x-6>0 \\ \Rightarrow 2x>6 \\ \Rightarrow x>3 \\ \Rightarrow x\in \left(3,\infty \right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(3,\infty \right). \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 2x-6<0 \\ \Rightarrow 2x<6 \\ \Rightarrow x<3 \\ \Rightarrow x\in \left(-\infty ,3\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(-\infty ,3\right)\text{.} \end{gathered}$$

Let (x, y) be the coordinates on the given curve where the normal to the curve is parallel to the given line.
Slope of the given line = 1

$$\begin{gathered} \text{Slope of tangent =}{\left(\frac{dy}{dx}\right)}_{\left(x,y\right)}\text{=}2x-6 \\ \text{Slope of normal =}\frac{-1}{\text{Slope of tangent}}\text{=}\frac{-1}{2x-6} \\ \mathrm{Now}, \\ \text{Slope of normal = Slope of the given line} \\ \frac{-1}{2x-6}=1 \\ -1=2x-6 \\ 2x=5 \\ x=\frac{5}{2} \\ \text{Given curve is} \\ y=x^2-6x+9 \\ =\frac{25}{4}-15+9 \\ =\frac{1}{4} \\ \left(x,y\right)=\left(\frac{5}{2},\frac{1}{4}\right) \\ \mathrm{Hence},\mathrm{the}\mathrm{coordinates}\mathrm{are}\left(\frac{5}{2},\frac{1}{4}\right). \end{gathered}$$