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.
Given:- Function f(x) = x2 – 6x + 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’(x) > 0 for all 
, then f(x) is increasing on (a, b)
(ii) If f’(x) < 0 for all 
, then f(x) is decreasing on (a, b)
Algorithm:-
(i) Obtain the function and put it equal to f(x)
(ii) Find f’(x)
(iii) Put f’(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) = x2 – 6x + 9
⇒ 
⇒ f’(x) = 2x – 6
⇒ f’(x) = 2(x – 3)
For f(x) lets find critical point, we must have
⇒ f’(x) = 0
⇒ 2(x – 3) = 0
⇒ (x – 3) = 0
⇒ x = 3
clearly, f’(x) > 0 if x > 3
and f’(x) < 0 if x < 3
Thus, f(x) increases on (3, ∞)
and f(x) is decreasing on interval x ∈ (–∞, 3)
Now, lets find coordinates of point
Equation of curve is
f(x) = x2 – 6x + 9
slope of this curve is given by
⇒ 
⇒ 
⇒ m1 = 2x – 6
and Equation of line is
y = x + 5
slope of this curve is given by
⇒ 
⇒ 
⇒ m2 = 1
Since slope of curve (i.e slope of its normal) is parallel to line
Therefore, they follow the relation
⇒ 
⇒ 
⇒ 2x – 6 = –1
⇒ 
Thus putting the value of x in equation of curve, we get
⇒ y = x2 – 6x + 9
⇒ 
⇒ 
⇒ 
⇒ 
Thus the required coordinates is 
)