Find the asymptotes of the following curve: f x = 3 x x 2 + 2

Given curve is f x = 3 x x 2 + 2 . For Vertical asymptote: Solve x 2 + 2 = 0 ⇒ x 2 + 2 = 0 x 2 = - 2 so, curve is always defined So there are no vertical asymptotes. For Horizontal asymptote: lim x → + ∞ 3 x x 2 + 2 = lim x → + ∞ 3 1 + 2 x 2 = ± 3 lim x → - ∞ 3 x x 2 + 2 = lim x → - ∞ 3 1 + 2 x 2 = ± 3 So y = 3 and y = - 3 are horizontal asymptotes. For Slant asymptote: The Polynomial in numerator does not have a higher degree than the polynomial in the denominator. So no slant asymptotes.

Find the asymptotes of the following curve: f x = x 2 - 6 x - 1 x + 3

Given f x = x 2 - 6 x - 1 x + 3 For Horizontal asymptote: lim x → + ∞ x 2 - 6 x - 1 x + 3 = lim x → + ∞ x - 6 x - 1 x 1 + 3 x = ∞ lim x → - ∞ x 2 - 6 x - 1 x + 3 = lim x → - ∞ x - 6 x - 1 x 1 + 3 x = - ∞ So no horizontal asymptotes. For Vertical asymptote: x + 3 = 0 ⇒ x = - 3 lim x → - 3 + x 2 - 6 x - 1 x + 3 = ∞ lim x → - 3 - x 2 - 6 x - 1 x + 3 = - ∞ So x = - 3 is a vertical asymptote. For Slant asymptote: The polynomial in the numerator is a higher degree than the polynomial in the denominator. To find the slant asymptote you must divide the numerator by the denominator using either long division or synthetic division. x + 3 x 2 - 6 x - 1 x - 9 x 2 + 3 x - 9 x - 1 - 9 x - 27 26 By performing synthetic division we get f x = x - 9 + 26 x + 3 So y = x - 9 is a slant asymptote. So, given curve has a vertical asymptote x = - 3 and slant asymptote y = x - 9 .

Find the asymptotes of the following curve: f x = x 2 + 6 x - 4 3 x - 6

Given f x = x 2 + 6 x - 4 3 x - 6 For Vertical asymptotes: 3 x - 6 = 0 ⇒ x = 2 , so at x = 2 lim x → 2 + x 2 + 6 x - 4 3 x - 6 = ∞ lim x → 2 - x 2 + 6 x - 4 3 x - 6 = - ∞ So x = 2 is a vertical asymptote. For Horizontal asymptotes: lim x → + ∞ x 2 + 6 x - 4 3 x - 6 = ∞ lim x → - ∞ x 2 + 6 x - 4 3 x - 6 = - ∞ So no Horizontal asymptotes. For Slant asymptotes: The polynomial in the numerator is a higher degree than the polynomial in the denominator. To find the slant asymptote you must divide the numerator by the denominator using either long division or synthetic division. 3 x - 6 x 2 + 6 x - 4 x 3 + 8 3 x 2 - 2 x 8 x - 4 8 x - 16 12 So, f x = x 3 + 8 3 + 12 3 x - 6 Then, y = x 3 + 8 3 is a Slant asymptote. So, given curve has a vertical asymptote x = 2 and slant asymptote y = x 3 + 8 3 .

Sketch the graph of the following function: y = - 1 3 x 3 - 3 x + 2

Given y = - 1 3 x 3 - 3 x + 2 Factorizing the given function, we have y = f x = - 1 3 x - 1 2 x + 2 The domain and the range of the given function f x are the entire real line. Putting y = 0 , we get x = 1 , - 2 . Therefore the x - intercepts are 1 , 0 and - 2 , 0 . putting x = 0 , we get y = - 2 3 . Therefore the y - intercept is 0 , - 2 3 . f ' x = - 1 3 3 x 2 - 3 and hence the critical points of the curve occur at x = ± 1 . f '' x = - 2 x . Therefore at x = 1 , the curve has a local maximum because f '' 1 = - 2 < 0 . The local maximum is f 1 = 0 . Similarly, at x = - 1 the curve has a local minimum because f '' - 1 = 2 > 0 . The local minimum is f - 1 = - 4 3 . Since f '' x = - 2 x < 0 ∀ x > 0 , the function is concave downward in the positive real line. As f '' x = - 2 x > 0 ∀ x < 0 , the function is concave upwards in the negative real line. Since f '' x = 0 at x = 0 and f '' x changes its sign when passing through x = 0 . Therefore, the point of inflection is 0 , f 0 = 0 , 2 3 . The curve has no asymptotes.

Sketch the graph of the following function: y = x 4 - x

Let f x = x 4 - x The domain of f x is x ≤ 4 . Putting y = 0 , we get x = 0 , 4 . Therefore the x - intercept is 4 , 0 . Putting x = 0 we get y = 0 . So the curve passes through the origin. f ' x = x - 1 2 4 - x + 4 - x = 8 - 3 x 2 4 - x and hence the critical point of the curve occurs at x = 4 as f ' x does not exist and the other critical point is x = 8 3 because f ' x = 0 . The monotone intervals are - ∞ < x < 8 3 , 8 3 < x < 4 . At x = 1 , f ' x > 0 and at x = 3 , f ' x < 0 . The Intervals of monotonocity is tabulated in the table below. - ∞ < x < 8 3 x = 8 3 8 3 < x < 4 x = 4 Sign of f ' x + 0 - 0 Behaviour Increasing Maximum Decreasing Minimum At x = 8 3 , the function has local maximum. The local maximum is 8 3 , f 8 3 = 8 3 , 16 3 3 . At x = 4 , the function has local minimum. The local minimum is 4 , f 4 = 4 , 0 . To determine the concavity, we find the second derivative as f " x = 21 x - 88 16 4 - x 3 2 . Therefore, when x < 8 3 , f " x < 0 . so the curve is concave downwards. When x ∈ 8 3 , 4 , f " x < 0 so the curve is concave downwards. Since f " x ≠ 0 , x ∈ - ∞ , 4 , there is no point of inflection. The curve has no asymptotes.

Sketch the graph of the following function: y = x 2 + 1 x 2 - 4

Let f x = x 2 + 1 x 2 - 4 = x 2 + 1 x + 2 x - 2 1. The domain of the function is - ∞ , - 2 ∪ - 2 , 2 ∪ 2 , ∞ . 2. Since - x , y = x , y the curve is symmetric about y - axis. 3. Putting y = 0 , we get x as imaginary points. So there are no x - intercepts. Putting x = 0 , we get y = - 1 4 . So the y - intercept is 0 , - 1 4 . 4. f ' x = - 10 x x 2 - 4 2 and hence the critical points of the curve occurs at x = 2 , - 2 as f ' x does not exist. The other critical point occurs at x = 0 as f ' x = 0 . The intervals of the monotonicity are tabulated below. - ∞ < x < - 2 - 2 < x < 0 0 < x < 2 2 < x < ∞ Sign + + - - Behaviour Increasing Increasing Decreasing Decreasing 5. f " x = 10 3 x 2 + 4 x 2 - 4 3 . At x = 0 , the curve has a local maximum because f " x = - 5 8 < 0 . The local maximum is 0 , f 0 = 0 , - 1 4 . As x = 2 , - 2 is not in the domain there is no local extrema at these points. The intervals of concavity are tabulated below: - ∞ , - 2 - 2 , 0 0 , 2 2 , ∞ Sign of f " x + - - + Concavity concave up concave down concave down concave up 6. As x = - 2 , 2 are not in the domain of f x and f " x ≠ 0 there is no point of inflection. 7. lim x → ± ∞ f x = lim x → ± ∞ x 2 + 1 x 2 - 4 = lim x → ± ∞ 1 + 1 x 2 1 - 4 x 2 = 1 . Therefore, y = 1 is a horizontal asymptote. 8. Since the denominator is zero when x = ± 2 . lim x → - 2 - x 2 + 1 x 2 - 4 = - ∞ , lim x → - 2 + x 2 + 1 x 2 - 4 = ∞ , lim x → 2 - x 2 + 1 x 2 - 4 = - ∞ , lim x → 2 + x 2 + 1 x 2 - 4 = ∞ Therefore, x = 2 , x = - 2 are vertical asymptotes.

Sketch the graph of the following function: y = 1 1 + e - x

Let f x = 1 1 + e - x 1. The Domain of f x is the entire real line and range is 0 , 1 . 2. Putting y = 0 , we get x as undefined, so there is no x - intercept. Putting x = 0 , we get y = 1 2 . So the y - intercept is 0 , 1 2 . 3. f ' x = e - x 1 + e - x 2 . So there are no critical points. Since there are no critical points, there are no extreme points. The function is constantly increasing because for a , b ∈ - ∞ , ∞ a ≤ b ⇒ f a ≤ f b . 4. f " x = e - x e - x - 1 e - x + 1 3 . So f " x > 0 for x > 0 . So the function is concave upward on - ∞ , 0 . f " x < 0 for x < 0 . So the function is concave downward on 0 , ∞ . 5. At x = 0 , f " x = 0 . So the point of inflection is 0 , f 0 = 0 , 1 2 . 6. lim x → + ∞ f x = lim x → + ∞ 1 1 + e - x = 1 lim x → - ∞ f x = lim x → - ∞ 1 1 + e - x = 0 . So, y = 0 , y = 1 are horizontal asymptotes.

Sketch the graph of the following function: y = x 3 24 - log x

Let f x = x 3 24 - log x 1. Domain of f x is 0 , ∞ . 2. Putting y = 0 , we get x undefined. So there are no x - intercepts. Putting x = 0 , we get y undefined. So there are no y - intercepts. 3. f ' x = x 2 8 - 1 x and hence the critical points of the curve occurs at x = 0 as f ' x does not exist. The other critical point occurs at x = 2 as f ' x = 0 . 4. f " x = x 4 + 1 x 2 , as x = 0 is not in the domain, there is no local extrema at this point. At x = 2 the function has local minimum because f " 2 = 3 4 > 0 . The local minimum is f 2 = 8 24 - log 2 = 1 3 - log 2 . 0 < x < 2 2 < x < ∞ Sign of f ' x - + Behaviour Decreasing Increasing 5. f " x > 0 ∀ x > 0 So the curve is concave upwards. 6. As f " x ≠ 0 , there are no points of inflection. 7. lim x → ∞ f x = lim x → ∞ x 3 24 - log x = ∞ and lim x → - ∞ f x = lim x → - ∞ x 3 24 - log x = - ∞ , so there are no horizontal asymptote. 8. Curve does not have vertical asymptote as denominator of the curve is never zero.

E M B I B E

Mathematics Standard 12 Vol II>Chapter 7 - Applications of Differential Calculus>Exercise 9

Tamil Nadu Board Solutions for Chapter: Applications of Differential Calculus, Exercise 9: Exercise 9

Author:Tamil Nadu Board

Tamil Nadu Board Mathematics Solutions for Exercise - Tamil Nadu Board Solutions for Chapter: Applications of Differential Calculus, Exercise 9: Exercise 9

Attempt the free practice questions on Chapter 7: Applications of Differential Calculus, Exercise 9: Exercise 9 with hints and solutions to strengthen your understanding. Mathematics Standard 12 Vol II solutions are prepared by Experienced Embibe Experts.