Samacheer Kalvi Class 12 Maths Solutions Chapter 7 Applications of Differential Calculus Exercise 7.7

Question 1. Find intervals of concavity and points of inflection for the following functions:
(i) \( f(x) = x(x – 4)^3 \)
(ii) \( f(x) = \sin x + \cos x, 0 < x < 2\pi \)
(iii) \( f(x) = \frac { 1 }{ 2 }(e^x – e^{-x}) \)
Answer:
(i) Given function is \( f(x) = x(x - 4)^3 \).
First, find the first derivative \( f'(x) \):
\( f'(x) = 1 \cdot (x - 4)^3 + x \cdot 3(x - 4)^2 \cdot 1 \)
\( f'(x) = (x - 4)^2 [ (x - 4) + 3x ] \)
\( f'(x) = (x - 4)^2 (4x - 4) = 4(x - 4)^2 (x - 1) \)
Next, find the second derivative \( f''(x) \):
\( f''(x) = 4 [ 2(x - 4) \cdot 1 \cdot (x - 1) + (x - 4)^2 \cdot 1 ] \)
\( f''(x) = 4 (x - 4) [ 2(x - 1) + (x - 4) ] \)
\( f''(x) = 4 (x - 4) [ 2x - 2 + x - 4 ] \)
\( f''(x) = 4 (x - 4) (3x - 6) \)
\( f''(x) = 12 (x - 4) (x - 2) \)
To find points of inflection, set \( f''(x) = 0 \):
\( 12 (x - 4) (x - 2) = 0 \)
\( \implies \) \( x - 4 = 0 \) or \( x - 2 = 0 \)
\( \implies \) \( x = 4 \) or \( x = 2 \)
These are the critical points for concavity. We examine the sign of \( f''(x) \) in the intervals defined by these points: \( (-\infty, 2) \), \( (2, 4) \), and \( (4, \infty) \).
- In \( (-\infty, 2) \), choose \( x = 0 \). \( f''(0) = 12(-4)(-2) = 96 > 0 \). So, the curve is concave upward.
- In \( (2, 4) \), choose \( x = 3 \). \( f''(3) = 12(-1)(1) = -12 < 0 \). So, the curve is concave downward.
- In \( (4, \infty) \), choose \( x = 5 \). \( f''(5) = 12(1)(3) = 36 > 0 \). So, the curve is concave upward.
The curve is concave upward in \( (-\infty, 2) \) and \( (4, \infty) \).
The curve is concave downward in \( (2, 4) \).
Points of inflection occur where \( f''(x) \) changes sign. This happens at \( x = 2 \) and \( x = 4 \).
- At \( x = 2 \): \( f(2) = 2(2 - 4)^3 = 2(-2)^3 = 2(-8) = -16 \).
- At \( x = 4 \): \( f(4) = 4(4 - 4)^3 = 4(0) = 0 \).
Thus, the points of inflection are \( (2, -16) \) and \( (4, 0) \). These are the points where the curve's concavity changes direction.
In simple words: To find where a graph bends (concave up or down) and where it changes its bend (inflection point), we take the second derivative. If it's positive, it bends up; if negative, it bends down. When it equals zero and changes sign, that's an inflection point.

(ii) Given function is \( f(x) = \sin x + \cos x \), for \( 0 < x < 2\pi \).
First, find the first derivative \( f'(x) \):
\( f'(x) = \cos x - \sin x \)
Next, find the second derivative \( f''(x) \):
\( f''(x) = -\sin x - \cos x \)
To find points of inflection, set \( f''(x) = 0 \):
\( -\sin x - \cos x = 0 \)
\( \implies \) \( \sin x + \cos x = 0 \)
\( \implies \) \( \tan x = -1 \)
For \( 0 < x < 2\pi \), the solutions for \( \tan x = -1 \) are:
\( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \)
These are the critical points for concavity. We examine the sign of \( f''(x) \) in the intervals defined by these points within the given domain \( (0, 2\pi) \): \( (0, \frac{3\pi}{4}) \), \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \), and \( (\frac{7\pi}{4}, 2\pi) \).
- In \( (0, \frac{3\pi}{4}) \), choose \( x = \frac{\pi}{2} \). \( f''(\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) - \cos(\frac{\pi}{2}) = -1 - 0 = -1 < 0 \). So, the curve is concave downward.
- In \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \), choose \( x = \pi \). \( f''(\pi) = -\sin(\pi) - \cos(\pi) = 0 - (-1) = 1 > 0 \). So, the curve is concave upward.
- In \( (\frac{7\pi}{4}, 2\pi) \), choose \( x = \frac{11\pi}{6} \). \( f''(\frac{11\pi}{6}) = -\sin(\frac{11\pi}{6}) - \cos(\frac{11\pi}{6}) = -(-\frac{1}{2}) - \frac{\sqrt{3}}{2} = \frac{1 - \sqrt{3}}{2} < 0 \). So, the curve is concave downward.
The curve is concave upward in \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \).
The curve is concave downward in \( (0, \frac{3\pi}{4}) \) and \( (\frac{7\pi}{4}, 2\pi) \).
Points of inflection occur where \( f''(x) \) changes sign. This happens at \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \).
- At \( x = \frac{3\pi}{4} \): \( f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{4}) + \cos(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0 \).
- At \( x = \frac{7\pi}{4} \): \( f(\frac{7\pi}{4}) = \sin(\frac{7\pi}{4}) + \cos(\frac{7\pi}{4}) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0 \).
Thus, the points of inflection are \( (\frac{3\pi}{4}, 0) \) and \( (\frac{7\pi}{4}, 0) \). These are specific points where the trigonometric function's curve changes its direction of bending.
In simple words: For trigonometric functions, we follow the same steps: find the second derivative and set it to zero to find potential inflection points. Then, check the sign of the second derivative in intervals to see if the curve is bending up or down.

(iii) Given function is \( f(x) = \frac{1}{2}(e^x - e^{-x}) \).
First, find the first derivative \( f'(x) \):
\( f'(x) = \frac{1}{2}(e^x - (-e^{-x})) = \frac{1}{2}(e^x + e^{-x}) \)
Next, find the second derivative \( f''(x) \):
\( f''(x) = \frac{1}{2}(e^x + (-e^{-x})) = \frac{1}{2}(e^x - e^{-x}) \)
To find points of inflection, set \( f''(x) = 0 \):
\( \frac{1}{2}(e^x - e^{-x}) = 0 \)
\( \implies \) \( e^x - e^{-x} = 0 \)
\( \implies \) \( e^x = e^{-x} \)
\( \implies \) \( e^x = \frac{1}{e^x} \)
\( \implies \) \( e^{2x} = 1 \)
\( \implies \) \( 2x = \ln(1) \)
\( \implies \) \( 2x = 0 \)
\( \implies \) \( x = 0 \)
This is the critical point for concavity. We examine the sign of \( f''(x) \) in the intervals \( (-\infty, 0) \) and \( (0, \infty) \).
- In \( (-\infty, 0) \), choose \( x = -1 \). \( f''(-1) = \frac{1}{2}(e^{-1} - e^{1}) = \frac{1}{2}(\frac{1}{e} - e) \). Since \( e \approx 2.718 \), \( \frac{1}{e} < e \), so \( \frac{1}{e} - e \) is negative. Thus, \( f''(-1) < 0 \). The curve is concave downward.
- In \( (0, \infty) \), choose \( x = 1 \). \( f''(1) = \frac{1}{2}(e^{1} - e^{-1}) = \frac{1}{2}(e - \frac{1}{e}) \). Since \( e > \frac{1}{e} \), \( e - \frac{1}{e} \) is positive. Thus, \( f''(1) > 0 \). The curve is concave upward.
The curve is concave upward in \( (0, \infty) \).
The curve is concave downward in \( (-\infty, 0) \).
Points of inflection occur where \( f''(x) \) changes sign. This happens at \( x = 0 \).
- At \( x = 0 \): \( f(0) = \frac{1}{2}(e^0 - e^{-0}) = \frac{1}{2}(1 - 1) = 0 \).
Thus, the point of inflection is \( (0, 0) \). This point is crucial as it indicates where the function's rate of change of slope itself changes.
In simple words: For functions involving 'e' (exponential functions), finding concavity and inflection points means calculating the second derivative and seeing when it becomes zero. Then, check the intervals to see if the graph is curving up or down.

🎯 Exam Tip: Remember to clearly state the intervals of concavity (upward and downward) and the exact coordinates of the inflection points. Always test a value in each interval to determine the sign of the second derivative.

Question 2. Find the local extrema for the following functions using second derivative test:
(i) \( f(x) = -3x^5 + 5x^3 \)
(ii) \( f(x) = x \log x \)
(iii) \( f(x) = x^2 e^{-2x} \)
Answer:
(i) Given function is \( f(x) = -3x^5 + 5x^3 \).
First, find the first derivative \( f'(x) \):
\( f'(x) = -15x^4 + 15x^2 \)
To find critical points, set \( f'(x) = 0 \):
\( -15x^4 + 15x^2 = 0 \)
\( \implies \) \( 15x^2(-x^2 + 1) = 0 \)
\( \implies \) \( 15x^2(1 - x)(1 + x) = 0 \)
\( \implies \) \( x = 0, x = 1, x = -1 \)
Next, find the second derivative \( f''(x) \):
\( f''(x) = -60x^3 + 30x \)
Now, apply the second derivative test at each critical point:
- At \( x = 0 \): \( f''(0) = -60(0)^3 + 30(0) = 0 \). The second derivative test is inconclusive here. In such cases, one would typically use the first derivative test, but following the source's flow, we'll proceed as shown. (Often, \(x=0\) in this function is an inflection point, not an extremum.)
- At \( x = 1 \): \( f''(1) = -60(1)^3 + 30(1) = -60 + 30 = -30 \). Since \( f''(1) < 0 \), there is a local maximum at \( x = 1 \).
The value of the function at \( x = 1 \) is \( f(1) = -3(1)^5 + 5(1)^3 = -3 + 5 = 2 \).
So, the local maximum point is \( (1, 2) \).
- At \( x = -1 \): \( f''(-1) = -60(-1)^3 + 30(-1) = -60(-1) - 30 = 60 - 30 = 30 \). Since \( f''(-1) > 0 \), there is a local minimum at \( x = -1 \).
The value of the function at \( x = -1 \) is \( f(-1) = -3(-1)^5 + 5(-1)^3 = -3(-1) + 5(-1) = 3 - 5 = -2 \).
So, the local minimum point is \( (-1, -2) \).
Therefore, the local maximum value is 2, and the local minimum value is -2. This test helps identify peaks and valleys on the graph.
In simple words: To find the highest and lowest points (local extrema) on a graph, we first find where the slope is zero using the first derivative. Then, we use the second derivative: if it's negative, it's a peak; if positive, it's a valley.

(ii) Given function is \( f(x) = x \log x \). (Assume natural logarithm, \( \ln x \)).
The domain of \( f(x) \) is \( x > 0 \).
First, find the first derivative \( f'(x) \):
\( f'(x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1 \)
To find critical points, set \( f'(x) = 0 \):
\( \log x + 1 = 0 \)
\( \implies \) \( \log x = -1 \)
\( \implies \) \( x = e^{-1} = \frac{1}{e} \)
Next, find the second derivative \( f''(x) \):
\( f''(x) = \frac{1}{x} \)
Now, apply the second derivative test at the critical point \( x = \frac{1}{e} \):
- At \( x = \frac{1}{e} \): \( f''(\frac{1}{e}) = \frac{1}{\frac{1}{e}} = e \). Since \( f''(\frac{1}{e}) = e > 0 \), there is a local minimum at \( x = \frac{1}{e} \).
The value of the function at \( x = \frac{1}{e} \) is \( f(\frac{1}{e}) = \frac{1}{e} \log(\frac{1}{e}) = \frac{1}{e} (-1) = -\frac{1}{e} \).
So, the local minimum point is \( (\frac{1}{e}, -\frac{1}{e}) \). This function has no local maximum, only a single lowest point in its domain.
In simple words: For this function, we find where its slope is flat. Then, we check the second derivative at that point. If it's positive, that spot is a low point on the graph.

(iii) Given function is \( f(x) = x^2 e^{-2x} \).
First, find the first derivative \( f'(x) \):
Using the product rule \( (uv)' = u'v + uv' \):
\( f'(x) = 2x \cdot e^{-2x} + x^2 \cdot (-2e^{-2x}) \)
\( f'(x) = 2x e^{-2x} - 2x^2 e^{-2x} \)
\( f'(x) = 2x e^{-2x} (1 - x) \)
To find critical points, set \( f'(x) = 0 \):
\( 2x e^{-2x} (1 - x) = 0 \)
Since \( e^{-2x} \) is never zero, we have:
\( 2x = 0 \) or \( 1 - x = 0 \)
\( \implies \) \( x = 0 \) or \( x = 1 \)
Next, find the second derivative \( f''(x) \):
We use the product rule again on \( f'(x) = (2x e^{-2x}) (1 - x) \). Let \( u = 2x e^{-2x} \) and \( v = (1 - x) \).
\( u' = 2e^{-2x} + 2x(-2e^{-2x}) = 2e^{-2x}(1 - 2x) \)
\( v' = -1 \)
\( f''(x) = u'v + uv' \)
\( f''(x) = [2e^{-2x}(1 - 2x)](1 - x) + [2x e^{-2x}](-1) \)
\( f''(x) = 2e^{-2x} [ (1 - 2x)(1 - x) - x ] \)
\( f''(x) = 2e^{-2x} [ 1 - x - 2x + 2x^2 - x ] \)
\( f''(x) = 2e^{-2x} [ 2x^2 - 4x + 1 ] \)
Now, apply the second derivative test at each critical point:
- At \( x = 0 \): \( f''(0) = 2e^0 [2(0)^2 - 4(0) + 1] = 2(1)[1] = 2 \). Since \( f''(0) > 0 \), there is a local minimum at \( x = 0 \).
The value of the function at \( x = 0 \) is \( f(0) = (0)^2 e^{-2(0)} = 0 \cdot 1 = 0 \).
So, the local minimum point is \( (0, 0) \).
- At \( x = 1 \): \( f''(1) = 2e^{-2(1)} [2(1)^2 - 4(1) + 1] = 2e^{-2} [2 - 4 + 1] = 2e^{-2} [-1] = -2e^{-2} \). Since \( f''(1) < 0 \), there is a local maximum at \( x = 1 \).
The value of the function at \( x = 1 \) is \( f(1) = (1)^2 e^{-2(1)} = 1 \cdot e^{-2} = e^{-2} = \frac{1}{e^2} \).
So, the local maximum point is \( (1, \frac{1}{e^2}) \). These points help define the overall shape and behavior of the function's graph.
In simple words: For this function, we find where the slope is zero. Then, we use the second derivative to see if these points are a low point (local minimum) or a high point (local maximum) on the graph.

🎯 Exam Tip: Be very careful with the chain rule and product rule when finding derivatives, especially for exponential functions. Ensure you test all critical points and correctly interpret the sign of the second derivative for maxima and minima.

Question 3. Find the intervals of monotonicity, local extrema, intervals of concavity and points of inflection for the function \( f(x) = 4x^3 + 3x^2 - 6x + 1 \).
Answer: Given function is \( f(x) = 4x^3 + 3x^2 - 6x + 1 \).

1. Monotonicity (Increasing/Decreasing Intervals) and Local Extrema:
First, find the first derivative \( f'(x) \):
\( f'(x) = 12x^2 + 6x - 6 \)
To find critical points, set \( f'(x) = 0 \):
\( 12x^2 + 6x - 6 = 0 \)
Divide by 6:
\( 2x^2 + x - 1 = 0 \)
Factor the quadratic:
\( (2x - 1)(x + 1) = 0 \)
\( \implies \) \( 2x - 1 = 0 \) or \( x + 1 = 0 \)
\( \implies \) \( x = \frac{1}{2} \) or \( x = -1 \)
These are the stationary points. Now, we examine the sign of \( f'(x) \) in the intervals defined by these points: \( (-\infty, -1) \), \( (-1, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \).
- In \( (-\infty, -1) \), choose \( x = -2 \). \( f'(-2) = 12(-2)^2 + 6(-2) - 6 = 12(4) - 12 - 6 = 48 - 18 = 30 > 0 \). So, \( f(x) \) is strictly increasing in \( (-\infty, -1) \).
- In \( (-1, \frac{1}{2}) \), choose \( x = 0 \). \( f'(0) = 12(0)^2 + 6(0) - 6 = -6 < 0 \). So, \( f(x) \) is strictly decreasing in \( (-1, \frac{1}{2}) \).
- In \( (\frac{1}{2}, \infty) \), choose \( x = 1 \). \( f'(1) = 12(1)^2 + 6(1) - 6 = 12 + 6 - 6 = 12 > 0 \). So, \( f(x) \) is strictly increasing in \( (\frac{1}{2}, \infty) \).

Local Extrema:
- At \( x = -1 \), \( f'(x) \) changes from positive to negative, indicating a local maximum.
\( f(-1) = 4(-1)^3 + 3(-1)^2 - 6(-1) + 1 = 4(-1) + 3(1) + 6 + 1 = -4 + 3 + 6 + 1 = 6 \).
Local maximum value is 6 at \( x = -1 \).
- At \( x = \frac{1}{2} \), \( f'(x) \) changes from negative to positive, indicating a local minimum.
\( f(\frac{1}{2}) = 4(\frac{1}{2})^3 + 3(\frac{1}{2})^2 - 6(\frac{1}{2}) + 1 \)
\( f(\frac{1}{2}) = 4(\frac{1}{8}) + 3(\frac{1}{4}) - 3 + 1 \)
\( f(\frac{1}{2}) = \frac{1}{2} + \frac{3}{4} - 2 = \frac{2}{4} + \frac{3}{4} - \frac{8}{4} = \frac{5 - 8}{4} = -\frac{3}{4} \).
Local minimum value is \( -\frac{3}{4} \) at \( x = \frac{1}{2} \).

2. Intervals of Concavity and Points of Inflection:
Next, find the second derivative \( f''(x) \):
\( f''(x) = \frac{d}{dx}(12x^2 + 6x - 6) = 24x + 6 \)
To find potential points of inflection, set \( f''(x) = 0 \):
\( 24x + 6 = 0 \)
\( \implies \) \( 24x = -6 \)
\( \implies \) \( x = -\frac{6}{24} = -\frac{1}{4} \)
This is a critical point for concavity. We examine the sign of \( f''(x) \) in the intervals defined by this point: \( (-\infty, -\frac{1}{4}) \) and \( (-\frac{1}{4}, \infty) \).
- In \( (-\infty, -\frac{1}{4}) \), choose \( x = -1 \). \( f''(-1) = 24(-1) + 6 = -24 + 6 = -18 < 0 \). So, the curve is concave downward.
- In \( (-\frac{1}{4}, \infty) \), choose \( x = 0 \). \( f''(0) = 24(0) + 6 = 6 > 0 \). So, the curve is concave upward.

Point of Inflection:
Since \( f''(x) \) changes sign at \( x = -\frac{1}{4} \), there is a point of inflection here.
To find the y-coordinate:
\( f(-\frac{1}{4}) = 4(-\frac{1}{4})^3 + 3(-\frac{1}{4})^2 - 6(-\frac{1}{4}) + 1 \)
\( f(-\frac{1}{4}) = 4(-\frac{1}{64}) + 3(\frac{1}{16}) + \frac{6}{4} + 1 \)
\( f(-\frac{1}{4}) = -\frac{1}{16} + \frac{3}{16} + \frac{3}{2} + 1 \)
\( f(-\frac{1}{4}) = \frac{2}{16} + \frac{24}{16} + \frac{16}{16} = \frac{2+24+16}{16} = \frac{42}{16} = \frac{21}{8} \).
Thus, the point of inflection is \( (-\frac{1}{4}, \frac{21}{8}) \). This point marks a transition in the curve's bending.
In simple words: We look at the first derivative to see where the function goes up or down and where its highest and lowest points are. Then, we use the second derivative to find where the curve changes how it bends, from curving up to curving down, or vice versa.

🎯 Exam Tip: For comprehensive analysis, always break down the problem into finding the first derivative (for monotonicity and extrema) and the second derivative (for concavity and inflection points). Clearly label your intervals and critical points.