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

\[\text { Here }, \]

\[f\left( x \right) = x^2 - 6x + 9\]

\[f'\left( x \right) = 2x - 6\]

\[\text { For f(x) to be increasing, we must have}\]

\[f'\left( x \right) > 0\]

\[ \Rightarrow 2x - 6 > 0\]

\[ \Rightarrow 2x > 6\]

\[ \Rightarrow x > 3\]

\[ \Rightarrow x \in \left( 3, \infty \right)\]

\[\text { So,f(x)is increasing on } \left( 3, \infty \right) . \]

\[\text { For }f(x) \text { to be decreasing, we must have }\]

\[f'\left( x \right) < 0\]

\[ \Rightarrow 2x - 6 < 0\]

\[ \Rightarrow 2x < 6\]

\[ \Rightarrow x < 3\]

\[ \Rightarrow x \in \left( - \infty , 3 \right)\]

\[\text { So,f(x)is decreasing on }\left( - \infty , 3 \right).\]

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

\[\text { Slope of tangent} = \left( \frac{dy}{dx} \right)_\left( x, y \right) =2x - 6\]

\[\text { Slope of normal } =\frac{- 1}{\text { Slope of tangent }}=\frac{- 1}{2x - 6}\]

\[\text { Now,} \]

\[\text { Slope of normal = Slope of the given line }\]

\[\frac{- 1}{2x - 6} = 1\]

\[ - 1 = 2x - 6\]

\[2x = 5\]

\[x = \frac{5}{2}\]

\[\text { Given curve is }\]

\[y = x^2 - 6x + 9\]

\[ = \frac{25}{4} - 15 + 9\]

\[ = \frac{1}{4}\]

\[\left( x, y \right) = \left( \frac{5}{2}, \frac{1}{4} \right)\]

\[\text { Hence, the coordinates are } \left( \frac{5}{2}, \frac{1}{4} \right) . \]