Question - 1 : -
Find the intervalsin which the following functions are increasing or decreasing.
(i) f (x) = 10 – 6x – 2x²
(ii) f (x) = x² + 2x – 5
(iii) f (x) = 6 –9x – x²
(iv) f(x) = 2x³ – 12x² +18x + 15
(v) f (x) = 5 + 36x+ 3x² – 2x³
(vi) f (x) = 8 + 36x + 3x_2 – 2x_3
(vii) f(x) = 5x_3 – 15x_2 – 120x + 3
(viii) f(x) = x_3 – 6x_2 – 36x + 2
(ix) f(x) = 2x_3 – 15x_2 + 36x + 1
(x) f (x) = 2x_3 + 9x_2 + 12x + 20

Answer - 1 : -

(i)

(ii)

(iii)

(iv)

(v)

Given f (x) = 5 + 36x + 3x² –2x³

⇒ f’(x)= 36 + 6x – 6x²

For f(x) now we have to find critical point, we must have

⇒ f’(x)= 0

⇒ 36+ 6x – 6x² = 0

⇒ 6(–x² +x + 6) = 0

⇒ 6(–x² +3x – 2x + 6) = 0

⇒ –x² +3x – 2x + 6 = 0

⇒ x² –3x + 2x – 6 = 0

⇒ (x– 3) (x + 2) = 0

⇒ x= 3, – 2

Clearly, f’(x) > 0 if –2< x < 3 and f’(x) < 0 if x< –2 and x > 3

Thus, f(x) increases on x ∈ (–2, 3) and f(x) is decreasing on interval (–∞, –2) ∪ (3, ∞)

(vi)

Given f (x) = 8 + 36x + 3x² – 2x³

Now differentiating with respect to x

⇒

⇒ f’(x)= 36 + 6x – 6x²

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

⇒ f’(x)= 0

⇒ 36+ 6x – 6x² = 0

⇒ 6(–x² +x + 6) = 0

⇒ 6(–x² +3x – 2x + 6) = 0

⇒ –x² +3x – 2x + 6 = 0

⇒ x² –3x + 2x – 6 = 0

⇒ (x– 3) (x + 2) = 0

⇒ x= 3, – 2

Clearly, f’(x) > 0 if –2 < x < 3 and f’(x) < 0 if x< –2 and x > 3

Thus, f(x) increases on x ∈ (–2, 3) and f(x) is decreasing on interval (–∞, 2) ∪ (3, ∞)

(vii)

Given f(x) = 5x³ – 15x² – 120x +3

Now by differentiating above equation with respect x, we get

⇒

⇒ f’(x)= 15x² – 30x – 120

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

⇒ f’(x)= 0

⇒ 15x² –30x – 120 = 0

⇒ 15(x² –2x – 8) = 0

⇒ 15(x² –4x + 2x – 8) = 0

⇒ x² –4x + 2x – 8 = 0

⇒ (x– 4) (x + 2) = 0

⇒ x= 4, – 2

Clearly, f’(x) > 0 if x < –2 and x > 4 and f’(x) < 0if –2 < x < 4

Thus, f(x) increases on (–∞,–2) ∪ (4, ∞) and f(x) is decreasing oninterval x ∈ (–2,4)

(viii)

Given f (x) = x³ – 6x² – 36x + 2

⇒

⇒ f’(x)= 3x² – 12x – 36

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

⇒ f’(x)= 0

⇒ 3x² –12x – 36 = 0

⇒ 3(x² –4x – 12) = 0

⇒ 3(x² –6x + 2x – 12) = 0

⇒ x² –6x + 2x – 12 = 0

⇒ (x– 6) (x + 2) = 0

⇒ x= 6, – 2

Clearly, f’(x) > 0 if x < –2 and x > 6 and f’(x) < 0if –2< x < 6

Thus, f(x) increases on (–∞,–2) ∪ (6, ∞) and f(x) is decreasing oninterval x ∈ (–2,6)

(ix)

Given f (x) = 2x³ – 15x² + 36x +1

Now by differentiating above equation with respect x, we get

⇒

⇒ f’(x)= 6x² – 30x + 36

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

⇒ f’(x)= 0

⇒ 6x² –30x + 36 = 0

⇒ 6(x² – 5x + 6) = 0

⇒ 6(x² –3x – 2x + 6) = 0

⇒ x² –3x – 2x + 6 = 0

⇒ (x– 3) (x – 2) = 0

⇒ x= 3, 2

Clearly, f’(x) > 0 if x < 2 and x > 3 and f’(x) < 0if 2 < x < 3

Thus, f(x) increases on (–∞, 2) ∪ (3, ∞) and f(x) is decreasing oninterval x ∈ (2,3)

(x)

Given f (x) = 2x³ + 9x² + 12x +20

Differentiating above equation we get

⇒

⇒ f’(x)= 6x² + 18x + 12

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

⇒ f’(x)= 0

⇒ 6x² +18x + 12 = 0

⇒ 6(x² +3x + 2) = 0

⇒ 6(x² +2x + x + 2) = 0

⇒ x² +2x + x + 2 = 0

⇒ (x+ 2) (x + 1) = 0

⇒ x= –1, –2

Clearly, f’(x) > 0 if –2 < x < –1 and f’(x) < 0 if x< –1 and x > –2

Thus, f(x) increases on x ∈ (–2,–1) and f(x) is decreasing on interval (–∞, –2) ∪ (–2, ∞)

Question - 2 : - 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 - 2 : -

Given f(x) = x² – 6x + 9

⇒

⇒ f’(x)= 2x – 6

⇒ f’(x)= 2(x – 3)

For f(x) let us 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 oninterval x ∈ (–∞,3)

Now, let us find coordinates of point

Equation of curve is f(x) = x² – 6x + 9

Slope of this curve is given by

Question - 3 : - Find the intervals in which f(x) = sin x – cos x, where 0 < x < 2π is increasing or decreasing.

Answer - 3 : -

Question - 4 : -
Show that f(x) = e^2x isincreasing on R.

Answer - 4 : -

Given f (x) = e^2x

⇒

⇒ f’(x)= 2e^2x

For f(x) to be increasing, we must have

⇒ f’(x)> 0

⇒ 2e^2x >0

⇒ e^2x >0

Since, the value of e lies between 2 and 3

So, whatever be the power of e (that is x in domain R) will begreater than zero.

Thus f(x) is increasing on interval R

Question - 5 : -
Show that f (x) = e^1/x,x ≠ 0 is a decreasing function for all x ≠ 0.

Answer - 5 : -

Question - 6 : -
Show that f(x) =log_a x, 0 < a < 1 is a decreasing function for all x >0.

Answer - 6 : -

Question - 7 : -
Show that f(x) =sin x is increasing on (0, π/2) and decreasing on (π/2, π) and neitherincreasing nor decreasing in (0, π).

Answer - 7 : -

Question - 8 : -
Show that f(x) =log sin x is increasing on (0, π/2) and decreasing on (π/2, π).

Answer - 8 : -

Question - 9 : -
Show that f(x) = x– sin x is increasing for all x ϵ R.

Answer - 9 : -

Given f (x) = x – sin x

⇒

⇒ f’(x)= 1 – cos x

Now, as given x ϵ R

⇒ –1< cos x < 1

⇒ –1> cos x > 0

⇒ f’(x)> 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ R

Question - 10 : -
Show that f(x) = x³ –15x² + 75x – 50 is an increasing function for allx ϵ R.

Answer - 10 : -

Given f(x) = x³ – 15x² + 75x –50

⇒

⇒ f’(x)= 3x² – 30x + 75

⇒ f’(x)= 3(x² – 10x + 25)

⇒ f’(x)= 3(x – 5)²

Now, as given x ϵ R

⇒ (x– 5)² > 0

⇒ 3(x– 5)² > 0

⇒ f’(x)> 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ R

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