## Book: RD Sharma - Mathematics (Volume 1)

### Chapter: 17. Increasing and Decreasing Functions

#### Subject: Maths - Class 12th

##### Q. No. 2 of Exercise 17.2

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##### 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 liney = 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

2x – 6 = –1

Thus putting the value of x in equation of curve, we get

y = x2 – 6x + 9

Thus the required coordinates is )

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