Samacheer Kalvi Class 12 Maths Solutions Chapter 7 Applications of Differential Calculus Exercise 7.9

Question 1. Find the asymptotes of the following curves:
(i) \( f(x) = \frac { x^2 }{ x^2-1} \)
(ii) \( f(x) = \frac { x^2 }{ x+1 } \)
(iii) \( f(x) = \frac { 3x }{ \sqrt{x^2+2} } \)
(iv) \( f(x) = \frac { x^2-6x-1 }{ x+3 } \)
(v) \( f(x) = \frac { x^2+6x-4 }{ 3x-6 } \)
Answer:
(i) For \( f(x) = \frac { x^2 }{ x^2-1} \):
To find vertical asymptotes, we set the denominator to zero:
\( x^2-1 = 0 \)
\( (x-1)(x+1) = 0 \)
\( \implies x = 1 \) and \( x = -1 \). These are the vertical asymptotes.
To find horizontal asymptotes, we look at the limit as \( x \rightarrow \pm \infty \):
\( \lim_{x \to \pm \infty} \frac{ x^2 }{ x^2-1} = \lim_{x \to \pm \infty} \frac { 1 }{ 1-\frac{1}{x^2} } = \frac { 1 }{ 1-0 } = 1 \)
\( \implies y = 1 \) is the horizontal asymptote. A function approaches its horizontal asymptote as \( x \) gets very large, either positive or negative.
In simple words: Vertical asymptotes happen where the bottom part of the fraction is zero. Horizontal asymptotes describe where the graph goes as x gets really big or really small.

🎯 Exam Tip: Remember to check both vertical (denominator zero) and horizontal (limit at infinity) asymptotes for rational functions. If the degree of the numerator is less than or equal to the degree of the denominator, there will be a horizontal asymptote.

Answer:
(ii) For \( f(x) = \frac { x^2 }{ x+1 } \):
To find vertical asymptotes, set the denominator to zero:
\( x+1 = 0 \)
\( \implies x = -1 \). This is the vertical asymptote.
To find horizontal asymptotes, we examine the limit as \( x \rightarrow \pm \infty \). Since the degree of the numerator (2) is greater than the degree of the denominator (1), there is no horizontal asymptote.
To find oblique asymptotes, we perform polynomial long division:
\[ \begin{array}{r} x - 1 \\ x+1 \overline{) x^2} \\ \underline{-(x^2+x)} \\ -x \\ \underline{-(-x-1)} \\ 1 \\ \end{array} \]
So, \( \frac{x^2}{x+1} = x - 1 + \frac{1}{x+1} \). As \( x \rightarrow \pm \infty \), \( \frac{1}{x+1} \rightarrow 0 \).
\( \implies y = x - 1 \) is the oblique (or slant) asymptote. Oblique asymptotes occur when the degree of the numerator is exactly one greater than the degree of the denominator.
In simple words: The graph has a vertical line where it breaks at x = -1. It doesn't flatten out horizontally, but instead, it gets closer and closer to a tilted line, which we find by dividing the top part by the bottom part.

🎯 Exam Tip: When the numerator's degree is one higher than the denominator's, always use polynomial long division to find the oblique asymptote.

Answer:
(iii) For \( f(x) = \frac { 3x }{ \sqrt{x^2+2} } \):
To find vertical asymptotes, we check if the denominator can be zero. Since \( x^2+2 \) is always positive (as \( x^2 \ge 0 \)), \( \sqrt{x^2+2} \) is never zero.
\( \implies \) There are no vertical asymptotes.
To find horizontal asymptotes, we evaluate limits at \( \pm \infty \):
For \( x \rightarrow \infty \):
\( \lim_{x \to \infty} \frac { 3x }{ \sqrt{x^2+2} } = \lim_{x \to \infty} \frac { 3x }{ \sqrt{x^2(1+\frac{2}{x^2})} } \)
\( = \lim_{x \to \infty} \frac { 3x }{ |x|\sqrt{1+\frac{2}{x^2}} } \)
Since \( x \rightarrow \infty \), \( |x| = x \).
\( = \lim_{x \to \infty} \frac { 3x }{ x\sqrt{1+\frac{2}{x^2}} } \)
\( = \lim_{x \to \infty} \frac { 3 }{ \sqrt{1+\frac{2}{x^2}} } = \frac { 3 }{ \sqrt{1+0} } = 3 \)
\( \implies y = 3 \) is a horizontal asymptote.
For \( x \rightarrow -\infty \):
\( \lim_{x \to -\infty} \frac { 3x }{ \sqrt{x^2+2} } = \lim_{x \to -\infty} \frac { 3x }{ |x|\sqrt{1+\frac{2}{x^2}} } \)
Since \( x \rightarrow -\infty \), \( |x| = -x \).
\( = \lim_{x \to -\infty} \frac { 3x }{ -x\sqrt{1+\frac{2}{x^2}} } \)
\( = \lim_{x \to -\infty} \frac { 3 }{ -\sqrt{1+\frac{2}{x^2}} } = \frac { 3 }{ -\sqrt{1+0} } = -3 \)
\( \implies y = -3 \) is another horizontal asymptote.
There are no slant asymptotes for this curve. This function has two distinct horizontal asymptotes as x approaches positive and negative infinity.
In simple words: There are no vertical breaks because the bottom part is never zero. As the graph goes far to the right, it gets close to the line y=3. As it goes far to the left, it gets close to the line y=-3.

🎯 Exam Tip: When dealing with square roots of \( x^2 \), remember that \( \sqrt{x^2} = |x| \). This means \( |x| = x \) for positive \( x \) and \( |x| = -x \) for negative \( x \), which can lead to different horizontal asymptotes.

Answer:
(iv) For \( f(x) = \frac { x^2-6x-1 }{ x+3 } \):
To find vertical asymptotes, set the denominator to zero:
\( x+3 = 0 \)
\( \implies x = -3 \). This is the vertical asymptote.
Since the degree of the numerator (2) is greater than the degree of the denominator (1), there is no horizontal asymptote.
To find oblique asymptotes, we perform polynomial long division:
\[ \begin{array}{r} x - 9 \\ x+3 \overline{) x^2-6x-1} \\ \underline{-(x^2+3x)} \\ -9x-1 \\ \underline{-(-9x-27)} \\ 26 \\ \end{array} \]
So, \( \frac{x^2-6x-1}{x+3} = x - 9 + \frac{26}{x+3} \). As \( x \rightarrow \pm \infty \), \( \frac{26}{x+3} \rightarrow 0 \).
\( \implies y = x - 9 \) is the oblique (or slant) asymptote. The remainder term disappears as x becomes very large.
In simple words: The graph has a vertical break at x = -3. It also follows a slanted line y = x-9 as it stretches out far away.

🎯 Exam Tip: Always state if horizontal asymptotes don't exist and proceed to check for oblique asymptotes using long division.

Answer:
(v) For \( f(x) = \frac { x^2+6x-4 }{ 3x-6 } \):
To find vertical asymptotes, set the denominator to zero:
\( 3x-6 = 0 \)
\( 3x = 6 \)
\( \implies x = 2 \). This is the vertical asymptote.
Since the degree of the numerator (2) is greater than the degree of the denominator (1), there is no horizontal asymptote.
To find oblique asymptotes, we perform polynomial long division:
\[ \begin{array}{r} \frac{x}{3} + \frac{8}{3} \\ 3x-6 \overline{) x^2+6x-4} \\ \underline{-(x^2-2x)} \\ 8x-4 \\ \underline{-(8x-16)} \\ 12 \\ \end{array} \]
So, \( \frac{x^2+6x-4}{3x-6} = \frac{x}{3} + \frac{8}{3} + \frac{12}{3x-6} \). As \( x \rightarrow \pm \infty \), \( \frac{12}{3x-6} \rightarrow 0 \).
\( \implies y = \frac{x}{3} + \frac{8}{3} \) (or \( 3y = x+8 \)) is the oblique (or slant) asymptote. The graph gets closer to this line as \( x \) moves far from the origin.
In simple words: The graph has a vertical break where x = 2. It does not have a horizontal line it approaches, but it does get very close to a slanted line as x gets very big or very small.

🎯 Exam Tip: When performing polynomial long division, be careful with signs when subtracting terms, as errors here are common.

Question 2. Sketch the graphs of the following functions
(i) \( y = -\frac { 1 }{ 3 } (x^3 - 3x + 2) \)
(ii) \( y = x \sqrt { 4-x } \)
(iii) \( y = \frac {x^2+1 }{ x^2-4 } \)
(iv) \( y = \frac { 1 }{ 1+e^{-x} } \)
(v) \( y = \frac { x^3 }{ 24 } - \log x \)
Answer:
(i) For \( y = -\frac { 1 }{ 3 } (x^3 - 3x + 2) \):
The domain and range for this polynomial function are all real numbers.
To find x-intercepts, set \( y = 0 \): \( x^3 - 3x + 2 = 0 \). Factoring gives \( (x-1)^2(x+2) = 0 \), so \( x = 1 \) (a double root) and \( x = -2 \). The x-intercepts are \( (1, 0) \) and \( (-2, 0) \).
To find the y-intercept, set \( x = 0 \): \( y = -\frac { 1 }{ 3 } (0 - 0 + 2) = -\frac { 2 }{ 3 } \). The y-intercept is \( (0, -\frac { 2 }{ 3 }) \).
First derivative: \( f'(x) = -\frac { 1 }{ 3 } (3x^2 - 3) = -(x^2 - 1) = 1 - x^2 \).
Critical points occur when \( f'(x) = 0 \): \( 1 - x^2 = 0 \implies x = \pm 1 \).
Second derivative: \( f''(x) = -2x \).
At \( x = 1 \): \( f''(1) = -2 < 0 \), so \( f(x) \) has a local maximum at \( x = 1 \), with \( f(1) = -\frac { 1 }{ 3 } (1 - 3 + 2) = 0 \).
At \( x = -1 \): \( f''(-1) = 2 > 0 \), so \( f(x) \) has a local minimum at \( x = -1 \), with \( f(-1) = -\frac { 1 }{ 3 } (-1 + 3 + 2) = -\frac { 4 }{ 3 } \).
Concavity: \( f''(x) = -2x \). For \( x > 0 \), \( f''(x) < 0 \), so the function is concave downward. For \( x < 0 \), \( f''(x) > 0 \), so the function is concave upward.
Point of inflection: \( f''(x) = 0 \) at \( x = 0 \). Since \( f''(x) \) changes sign at \( x = 0 \), \( (0, -\frac { 2 }{ 3 }) \) is the point of inflection. This point shows where the curve changes its bending direction.
This curve, being a polynomial, has no asymptotes.
In simple words: This graph goes through all x and y values. It crosses the x-axis at -2 and 1, and the y-axis at -2/3. It has a high point at (1,0) and a low point at (-1, -4/3). It bends downwards for positive x and upwards for negative x, changing its bend at (0, -2/3).

🎯 Exam Tip: For polynomial functions, always look for real roots to find x-intercepts, and calculate both first and second derivatives to determine local extrema and concavity.

Answer:
(ii) For \( y = x \sqrt { 4-x } \):
The expression \( \sqrt{4-x} \) requires \( 4-x \ge 0 \), so \( x \le 4 \). Thus, the domain is \( (-\infty, 4] \). The curve does not exist for \( x > 4 \).
The range is \( (-\infty, \frac { 16 }{ 3\sqrt{3} }] \).
To find x-intercepts, set \( y = 0 \): \( x \sqrt { 4-x } = 0 \implies x = 0 \) or \( 4-x = 0 \implies x = 4 \). The x-intercepts are \( (0, 0) \) and \( (4, 0) \). The curve passes through the origin.
To find y-intercept, set \( x = 0 \): \( y = 0 \sqrt { 4-0 } = 0 \). The y-intercept is \( (0, 0) \).
First derivative: \( f'(x) = 1 \cdot \sqrt{4-x} + x \cdot \frac{1}{2\sqrt{4-x}} \cdot (-1) = \sqrt{4-x} - \frac{x}{2\sqrt{4-x}} \)
\( = \frac{2(4-x) - x}{2\sqrt{4-x}} = \frac{8-2x-x}{2\sqrt{4-x}} = \frac{8-3x}{2\sqrt{4-x}} \).
Critical points occur when \( f'(x) = 0 \): \( 8-3x = 0 \implies x = \frac{8}{3} \).
Second derivative: \( f''(x) = \frac{-3(2\sqrt{4-x}) - (8-3x)(\frac{-1}{\sqrt{4-x}})}{4(4-x)} = \frac{-6(4-x) + (8-3x)}{4(4-x)^{3/2}} = \frac{-24+6x+8-3x}{4(4-x)^{3/2}} = \frac{3x-16}{4(4-x)^{3/2}} \).
At \( x = \frac{8}{3} \): \( f''(\frac{8}{3}) = \frac{3(\frac{8}{3})-16}{4(4-\frac{8}{3})^{3/2}} = \frac{8-16}{4(\frac{4}{3})^{3/2}} = \frac{-8}{4(\frac{8}{3\sqrt{3}})} = -\frac{8 \cdot 3\sqrt{3}}{32} = -\frac{3\sqrt{3}}{4} < 0 \).
So, \( f(x) \) has a local maximum at \( x = \frac{8}{3} \). The local maximum value is \( f(\frac{8}{3}) = \frac{8}{3} \sqrt{4-\frac{8}{3}} = \frac{8}{3} \sqrt{\frac{4}{3}} = \frac{8}{3} \frac{2}{\sqrt{3}} = \frac{16}{3\sqrt{3}} \).
At \( x=4 \), \( f(4)=0 \), which is a local minimum from the graph.
Concavity: \( f''(x) = \frac{3x-16}{4(4-x)^{3/2}} \). For \( x < 4 \) and \( 3x-16 < 0 \) (i.e., \( x < \frac{16}{3} \)), \( f''(x) < 0 \), so the curve is concave downward in its domain. \( \frac{16}{3} \) is outside the domain, so it's always concave downward within its domain. This means the curve has no point of inflection.
As \( x \rightarrow -\infty \), \( y = x \sqrt { 4-x } \rightarrow (-\infty) \cdot (\infty) = -\infty \). The curve goes downwards without any horizontal or oblique asymptotes.
In simple words: The graph only exists for x values up to 4. It starts from the left, goes up to a high point, then comes back down to touch the x-axis at 4. It has no straight lines it gets close to at its ends.

🎯 Exam Tip: Always pay attention to the domain when square roots are involved, as it significantly impacts the graph's existence and behavior.

Answer:
(iii) For \( y = \frac {x^2+1 }{ x^2-4 } \):
The function is undefined when the denominator is zero: \( x^2-4 = 0 \implies (x-2)(x+2) = 0 \).
\( \implies x = 2 \) and \( x = -2 \) are vertical asymptotes. The domain is \( (-\infty, -2) \cup (-2, 2) \cup (2, \infty) \).
To find the x-intercepts, set \( y = 0 \): \( x^2+1 = 0 \implies x^2 = -1 \). There are no real solutions, so no x-intercepts.
To find the y-intercept, set \( x = 0 \): \( y = \frac{0^2+1}{0^2-4} = -\frac{1}{4} \). The y-intercept is \( (0, -\frac{1}{4}) \).
To find horizontal asymptotes:
\( \lim_{x \to \pm \infty} \frac{x^2+1}{x^2-4} = \lim_{x \to \pm \infty} \frac{1+\frac{1}{x^2}}{1-\frac{4}{x^2}} = \frac{1+0}{1-0} = 1 \).
\( \implies y = 1 \) is the horizontal asymptote. The range of the function is \( (-\infty, -\frac{1}{4}] \cup (1, \infty) \).
First derivative: \( f'(x) = \frac{2x(x^2-4) - (x^2+1)(2x)}{(x^2-4)^2} = \frac{2x^3-8x-2x^3-2x}{(x^2-4)^2} = \frac{-10x}{(x^2-4)^2} \).
Critical points: \( f'(x) = 0 \implies -10x = 0 \implies x = 0 \).
Second derivative: \( f''(x) = \frac{-10(x^2-4)^2 - (-10x) \cdot 2(x^2-4)(2x)}{(x^2-4)^4} = \frac{-10(x^2-4) + 40x^2}{(x^2-4)^3} = \frac{-10x^2+40+40x^2}{(x^2-4)^3} = \frac{30x^2+40}{(x^2-4)^3} \).
At \( x = 0 \): \( f''(0) = \frac{40}{(-4)^3} = \frac{40}{-64} < 0 \). So, \( f(x) \) has a local maximum at \( x = 0 \). The local maximum value is \( f(0) = -\frac{1}{4} \).
Points of inflection: \( f''(x) = 0 \implies 30x^2+40 = 0 \). No real solutions, so no points of inflection.
In simple words: The graph breaks vertically at x = -2 and x = 2. It never crosses the x-axis. It crosses the y-axis at -1/4. As x gets very large, the graph flattens out towards y = 1. It has a high point at (0, -1/4).

🎯 Exam Tip: Remember that no x-intercepts mean the curve never crosses the x-axis, which is often characteristic of rational functions with positive numerators like \( x^2+1 \).

Answer:
(iv) For \( y = \frac { 1 }{ 1+e^{-x} } \):
The domain of this function is all real numbers, \( (-\infty, \infty) \).
To find x-intercepts, set \( y = 0 \): \( \frac { 1 }{ 1+e^{-x} } = 0 \). This has no solution since the numerator is always 1. Thus, there are no x-intercepts.
To find the y-intercept, set \( x = 0 \): \( y = \frac { 1 }{ 1+e^{0} } = \frac { 1 }{ 1+1 } = \frac { 1 }{ 2 } \). The y-intercept is \( (0, \frac{1}{2}) \).
First derivative: \( f'(x) = \frac{d}{dx} (1+e^{-x})^{-1} = -(1+e^{-x})^{-2}(-e^{-x}) = \frac{e^{-x}}{(1+e^{-x})^2} \).
Critical points: \( f'(x) = 0 \implies e^{-x} = 0 \). This is absurd, as \( e^{-x} \) is always positive. So, there are no extremum points (maxima or minima).
Asymptotes:
Vertical asymptotes: The denominator \( 1+e^{-x} \) is never zero (since \( e^{-x} > 0 \)). So, there are no vertical asymptotes.
Horizontal asymptotes:
As \( x \rightarrow \infty \): \( \lim_{x \to \infty} \frac { 1 }{ 1+e^{-x} } = \frac { 1 }{ 1+0 } = 1 \). So, \( y = 1 \) is a horizontal asymptote.
As \( x \rightarrow -\infty \): \( \lim_{x \to -\infty} \frac { 1 }{ 1+e^{-x} } = \frac { 1 }{ 1+\infty } = 0 \). So, \( y = 0 \) is another horizontal asymptote.
The range of the function is \( (0, 1) \), meaning \( 0 < f(x) < 1 \). This function is commonly known as the logistic function, which shows steady growth towards a limit.
In simple words: This graph exists for all numbers. It never touches the x-axis, but it does cross the y-axis at 1/2. As you go far to the right, the graph flattens towards y=1. As you go far to the left, it flattens towards y=0. It always goes upwards and has no high or low points.

🎯 Exam Tip: Functions involving exponentials like \( e^{-x} \) often have horizontal asymptotes because \( e^{-x} \) approaches 0 as \( x \rightarrow \infty \) and becomes very large as \( x \rightarrow -\infty \).

Answer:
(v) For \( y = \frac { x^3 }{ 24 } - \log x \):
The logarithm function \( \log x \) is only defined for positive values of \( x \). So, the domain of the curve is \( (0, \infty) \).
There are no x-intercepts because \( \frac { x^3 }{ 24 } = \log x \) has no real solutions, and no y-intercept since \( x \) cannot be 0.
Asymptotes:
Vertical asymptote: As \( x \rightarrow 0^+ \), \( \log x \rightarrow -\infty \). So \( y = \frac { x^3 }{ 24 } - \log x \rightarrow 0 - (-\infty) = \infty \).
\( \implies x = 0 \) (the y-axis) is a vertical asymptote. This means the graph gets very close to the y-axis but never touches or crosses it.
Horizontal asymptotes: As \( x \rightarrow \infty \), \( \frac{x^3}{24} \rightarrow \infty \) and \( \log x \rightarrow \infty \). The cubic term grows much faster than the logarithm, so \( y \rightarrow \infty \). There are no horizontal asymptotes.
First derivative: \( f'(x) = \frac{3x^2}{24} - \frac{1}{x} = \frac{x^2}{8} - \frac{1}{x} \).
Critical points: \( f'(x) = 0 \implies \frac{x^2}{8} = \frac{1}{x} \implies x^3 = 8 \implies x = 2 \).
Second derivative: \( f''(x) = \frac{2x}{8} - (-\frac{1}{x^2}) = \frac{x}{4} + \frac{1}{x^2} \).
At \( x = 2 \): \( f''(2) = \frac{2}{4} + \frac{1}{2^2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} > 0 \).
So, \( f(x) \) has a local minimum at \( x = 2 \). The local minimum value is \( f(2) = \frac{2^3}{24} - \log 2 = \frac{8}{24} - \log 2 = \frac{1}{3} - \log 2 \).
Points of inflection: \( f''(x) = 0 \implies \frac{x}{4} + \frac{1}{x^2} = 0 \implies x^3+4 = 0 \implies x = \sqrt[3]{-4} \). This is not in the domain \( (0, \infty) \), so there are no points of inflection.
The range of the function is \( [\frac{1}{3} - \log 2, \infty) \).
In simple words: This graph only exists for positive x values. It doesn't cross the y-axis, but it gets very close to it as x gets tiny. As x gets very big, the graph goes upwards forever. It has a lowest point when x is 2.

🎯 Exam Tip: When a function includes \( \log x \), always start by identifying its domain \( (x > 0) \), which instantly tells you there will be a vertical asymptote at \( x=0 \).