OP Malhotra Class 12 Maths Solutions Chapter 12 Maxima and Minima Exercise 12 (A)

S Chand Class 12 ICSE Maths Solutions Chapter 12 Maxima and Minima Ex 12(a)

Question 1. Find the turning values of the following functions, distinguishing in each case whether the value is a maximum, minimum, or inflexional:
(i) \( 4x^3 + 19x^2 - 14x + 3 \)
(ii) \( 2x^3 + 3x^2 - 12x + 7 \)
(iii) \( 3x^4 + 8x^3 + 6x^2 \)
Answer:
(i) Let the given function be \( y = 4x^3 + 19x^2 - 14x + 3 \).
First, we find the first derivative of y with respect to x:
\( \frac { dy }{ dx } = 12x^2 + 38x - 14 \)
Next, we find the second derivative of y:
\( \frac{d^2 y}{d x^2} = 24x + 38 \)
To find the points where maxima or minima occur, we set the first derivative to zero:
\( \frac { dy }{ dx } = 0 \)
\( \implies \) \( 12x^2 + 38x - 14 = 0 \)
We can simplify this by dividing by 2: \( 6x^2 + 19x - 7 = 0 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \):
\( x = \frac{-19 \pm \sqrt{19^2 - 4(6)(-7)}}{2(6)} \)
\( x = \frac{-19 \pm \sqrt{361 + 168}}{12} \)
\( x = \frac{-19 \pm \sqrt{529}}{12} \)
\( x = \frac{-19 \pm 23}{12} \)
This gives us two possible values for x:
\( x = \frac{-19 + 23}{12} = \frac{4}{12} = \frac{1}{3} \)
or
\( x = \frac{-19 - 23}{12} = \frac{-42}{12} = -\frac{7}{2} \)

Now, we check the second derivative at these points to determine if they are maxima or minima.
When \( x = \frac{1}{3} \):
\( \frac{d^2 y}{d x^2} = 24 \left( \frac{1}{3} \right) + 38 = 8 + 38 = 46 \)
Since \( 46 > 0 \), \( x = \frac{1}{3} \) is a point of local minima. The function curves upwards at this point.
The minimum value of y is:
\( y = 4\left(\frac{1}{3}\right)^3 + 19\left(\frac{1}{3}\right)^2 - 14\left(\frac{1}{3}\right) + 3 \)
\( y = 4\left(\frac{1}{27}\right) + 19\left(\frac{1}{9}\right) - \frac{14}{3} + 3 \)
\( y = \frac{4}{27} + \frac{19}{9} - \frac{14}{3} + 3 \)
To sum these fractions, we use a common denominator of 27:
\( y = \frac{4}{27} + \frac{19 \times 3}{9 \times 3} - \frac{14 \times 9}{3 \times 9} + \frac{3 \times 27}{1 \times 27} \)
\( y = \frac{4 + 57 - 126 + 81}{27} = \frac{16}{27} \)

When \( x = -\frac{7}{2} \):
\( \frac{d^2 y}{d x^2} = 24 \left( -\frac{7}{2} \right) + 38 = -12 \times 7 + 38 = -84 + 38 = -46 \)
Since \( -46 < 0 \), \( x = -\frac{7}{2} \) is a point of local maxima. The function curves downwards at this point.
The maximum value of y is:
\( y = 4\left(-\frac{7}{2}\right)^3 + 19\left(-\frac{7}{2}\right)^2 - 14\left(-\frac{7}{2}\right) + 3 \)
\( y = 4\left(-\frac{343}{8}\right) + 19\left(\frac{49}{4}\right) + 49 + 3 \)
\( y = -\frac{343}{2} + \frac{931}{4} + 49 + 3 \)
To sum these terms, we use a common denominator of 4:
\( y = \frac{-343 \times 2}{2 \times 2} + \frac{931}{4} + \frac{49 \times 4}{1 \times 4} + \frac{3 \times 4}{1 \times 4} \)
\( y = \frac{-686 + 931 + 196 + 12}{4} = \frac{453}{4} \)
We can also write this as a mixed number: \( 113\frac{1}{4} \).
In simple words: To find where a curve turns, we use calculus. First, we find the critical points by taking the first derivative and setting it to zero. Then, we use the second derivative to check if these points are high points (maxima) or low points (minima) on the graph.

(ii) Let the given function be \( y = 2x^3 + 3x^2 - 12x + 7 \).
First, we find the first derivative:
\( \frac{dy}{dx} = 6x^2 + 6x - 12 \)
Next, we find the second derivative:
\( \frac{d^2 y}{d x^2} = 12x + 6 \)
To find critical points, we set the first derivative to zero:
\( \frac{dy}{dx} = 0 \)
\( \implies \) \( 6x^2 + 6x - 12 = 0 \)
Divide by 6: \( x^2 + x - 2 = 0 \)
Factor the quadratic equation:
\( (x + 2)(x - 1) = 0 \)
This gives us two values for x: \( x = 1 \) or \( x = -2 \).

Now, we check the second derivative at these points.
When \( x = 1 \):
\( \left( \frac{d^2 y}{d x^2} \right)_{x=1} = 12(1) + 6 = 18 \)
Since \( 18 > 0 \), \( x = 1 \) is a point of local minima.
The minimum value of y is:
\( y = 2(1)^3 + 3(1)^2 - 12(1) + 7 \)
\( y = 2 + 3 - 12 + 7 = 0 \)

When \( x = -2 \):
\( \left( \frac{d^2 y}{d x^2} \right)_{x=-2} = 12(-2) + 6 = -24 + 6 = -18 \)
Since \( -18 < 0 \), \( x = -2 \) is a point of local maxima.
The maximum value of y is:
\( y = 2(-2)^3 + 3(-2)^2 - 12(-2) + 7 \)
\( y = 2(-8) + 3(4) + 24 + 7 \)
\( y = -16 + 12 + 24 + 7 = 27 \)

(iii) Let the given function be \( y = 3x^4 + 8x^3 + 6x^2 \).
First, we find the first derivative:
\( \frac{dy}{dx} = 12x^3 + 24x^2 + 12x \)
Next, we find the second derivative:
\( \frac{d^2 y}{d x^2} = 36x^2 + 48x + 12 \)
To find critical points, we set the first derivative to zero:
\( \frac{dy}{dx} = 0 \)
\( \implies \) \( 12x^3 + 24x^2 + 12x = 0 \)
Factor out \( 12x \): \( 12x(x^2 + 2x + 1) = 0 \)
Recognize the perfect square: \( 12x(x + 1)^2 = 0 \)
This gives us two critical values for x: \( x = 0 \) or \( x = -1 \).

Now, we check the second derivative at these points.
When \( x = 0 \):
\( \left( \frac{d^2 y}{d x^2} \right)_{x=0} = 36(0)^2 + 48(0) + 12 = 12 \)
Since \( 12 > 0 \), \( x = 0 \) is a point of local minima.
The minimum value of y is:
\( y = 3(0)^4 + 8(0)^3 + 6(0)^2 = 0 \)

When \( x = -1 \):
\( \left( \frac{d^2 y}{d x^2} \right)_{x=-1} = 36(-1)^2 + 48(-1) + 12 = 36 - 48 + 12 = 0 \)
Since the second derivative is zero, we need to check the third derivative to determine the nature of the point. If the second derivative is zero, it could be a point of inflexion.
Let's find the third derivative:
\( \frac{d^3 y}{d x^3} = 72x + 48 \)
Now, we evaluate the third derivative at \( x = -1 \):
\( \left( \frac{d^3 y}{d x^3} \right)_{x=-1} = 72(-1) + 48 = -72 + 48 = -24 \)
Since \( -24 \neq 0 \), \( x = -1 \) is a point of inflexion. At this point, the curve changes its concavity.
In simple words: We find the turning points by setting the first derivative to zero. Then, we use the second derivative to tell if it's a lowest point (minimum) or highest point (maximum). If the second derivative is also zero, we look at the third derivative to see if it's an inflexion point, which means the curve changes how it bends.

🎯 Exam Tip: Remember to calculate the actual values of y at the maxima and minima points. If the second derivative is zero, always proceed to calculate the third derivative to confirm if it's a point of inflexion.

Question 2. If \( V = 2x^2 (6 - x) \), where x is positive, determine the greatest value of V.
Answer: Let the given function be \( V = 2x^2(6 - x) \).
First, expand the function: \( V = 12x^2 - 2x^3 \).
Now, we find the first derivative of V with respect to x:
\( \frac{dV}{dx} = 24x - 6x^2 \)
We can factor this as \( \frac{dV}{dx} = 6x(4 - x) \).
Next, we find the second derivative of V:
\( \frac{d^2 V}{d x^2} = 24 - 12x \)
To find the points for maxima or minima, we set the first derivative to zero:
\( \frac{dV}{dx} = 0 \)
\( \implies \) \( 6x(4 - x) = 0 \)
This gives us two critical values for x: \( x = 0 \) or \( x = 4 \).

Since the problem states that x is positive, we consider \( x = 4 \). However, we will also check \( x = 0 \) for completeness.
When \( x = 0 \):
\( \left( \frac{d^2 V}{d x^2} \right)_{x=0} = 24 - 12(0) = 24 \)
Since \( 24 > 0 \), \( x = 0 \) is a point of local minima, and the minimum value of V is \( 2(0)^2(6 - 0) = 0 \).

When \( x = 4 \):
\( \left( \frac{d^2 V}{d x^2} \right)_{x=4} = 24 - 12(4) = 24 - 48 = -24 \)
Since \( -24 < 0 \), \( x = 4 \) is a point of local maxima. This means the value of V is greatest at this point.
The greatest value of V is:
\( V = 2(4)^2(6 - 4) \)
\( V = 2(16)(2) \)
\( V = 64 \)
In simple words: To find the highest value of a function, we take its first derivative and find where it equals zero. Then, we use the second derivative to confirm if it's a peak point. Here, the function V reaches its highest point of 64 when x is 4.

🎯 Exam Tip: When finding greatest or least values, always ensure you use the specified domain for x (e.g., positive x) and check the nature of critical points using the second derivative test.

Question 3. Find the co-ordinates of the turning points on the curve \( y = x^3 - 3x^2 - 9x + 7 \), distinguishing between maximum and minimum points.
Answer: Let the given curve be \( y = x^3 - 3x^2 - 9x + 7 \).
First, we find the first derivative of y with respect to x:
\( \frac{dy}{dx} = 3x^2 - 6x - 9 \)
Next, we find the second derivative:
\( \frac{d^2 y}{d x^2} = 6x - 6 \)
To find the turning points, we set the first derivative to zero:
\( \frac{dy}{dx} = 0 \)
\( \implies \) \( 3x^2 - 6x - 9 = 0 \)
Divide by 3: \( x^2 - 2x - 3 = 0 \)
Factor the quadratic equation:
\( (x + 1)(x - 3) = 0 \)
This gives us two critical values for x: \( x = -1 \) or \( x = 3 \).

Now, we check the second derivative at these points to identify maxima or minima.
When \( x = -1 \):
\( \left( \frac{d^2 y}{d x^2} \right)_{x=-1} = 6(-1) - 6 = -6 - 6 = -12 \)
Since \( -12 < 0 \), \( x = -1 \) is a point of local maxima.
To find the maximum value of y, substitute \( x = -1 \) into the original equation:
\( y = (-1)^3 - 3(-1)^2 - 9(-1) + 7 \)
\( y = -1 - 3(1) + 9 + 7 \)
\( y = -1 - 3 + 9 + 7 = 12 \)
So, the maximum point is \( (-1, 12) \).

When \( x = 3 \):
\( \left( \frac{d^2 y}{d x^2} \right)_{x=3} = 6(3) - 6 = 18 - 6 = 12 \)
Since \( 12 > 0 \), \( x = 3 \) is a point of local minima.
To find the minimum value of y, substitute \( x = 3 \) into the original equation:
\( y = (3)^3 - 3(3)^2 - 9(3) + 7 \)
\( y = 27 - 3(9) - 27 + 7 \)
\( y = 27 - 27 - 27 + 7 = -20 \)
So, the minimum point is \( (3, -20) \).
The turning points are \( (-1, 12) \) and \( (3, -20) \).
In simple words: Turning points on a curve are where its slope becomes flat. We find these points by setting the first derivative to zero. Then, we use the second derivative to see if the curve is at a peak (maximum) or a valley (minimum) at these points.

🎯 Exam Tip: Always calculate both the x and y coordinates of the turning points to provide a complete answer for "co-ordinates". Remember to substitute x back into the original function for y.

Question 4. Find the minimum value of \( \left(x+\frac{4}{x^2}\right) \).
Answer: Let the given function be \( y = x + \frac{4}{x^2} \).
We can rewrite this as \( y = x + 4x^{-2} \).
First, we find the first derivative:
\( \frac{dy}{dx} = 1 - 8x^{-3} = 1 - \frac{8}{x^3} \)
Next, we find the second derivative:
\( \frac{d^2 y}{d x^2} = -8(-3)x^{-4} = 24x^{-4} = \frac{24}{x^4} \)
To find critical points, we set the first derivative to zero:
\( \frac{dy}{dx} = 0 \)
\( \implies \) \( 1 - \frac{8}{x^3} = 0 \)
\( \implies \) \( 1 = \frac{8}{x^3} \)
\( \implies \) \( x^3 = 8 \)
\( \implies \) \( x = 2 \)
The equation \( x^3 - 8 = 0 \) can be factored as \( (x - 2)(x^2 + 2x + 4) = 0 \). The quadratic part \( x^2 + 2x + 4 \) has no real roots, so \( x = 2 \) is the only real critical point.

Now, we check the second derivative at \( x = 2 \):
\( \left( \frac{d^2 y}{d x^2} \right)_{x=2} = \frac{24}{2^4} = \frac{24}{16} = \frac{3}{2} \)
Since \( \frac{3}{2} > 0 \), \( x = 2 \) is a point of local minima.
The minimum value of y is:
\( y = 2 + \frac{4}{2^2} \)
\( y = 2 + \frac{4}{4} \)
\( y = 2 + 1 = 3 \)
In simple words: To find the smallest value of this function, we calculate its first derivative and find where it is zero. This gives us the point where the curve flattens. Then, we use the second derivative to confirm that this point is indeed the lowest point on the curve, and we find the function's value at that point.

🎯 Exam Tip: Always analyze the cubic equation for its roots carefully. When using the second derivative test, ensure that you only consider real roots for finding maxima or minima.

Question 5. If \( y = \frac { 1 }{ 4 }x^4 - \frac { 2 }{ 3 }x^3 + \frac { 1 }{ 2 }x^2 + \frac { 11 }{ 2 } \), show that the ordinate at the point x = 1 is neither a maximum nor a minimum, though \( \frac{dy}{dx} = 0 \), when x = 1.
Answer: Let the given function be \( y = \frac{x^4}{4} - \frac{2}{3}x^3 + \frac{x^2}{2} + \frac{11}{2} \).
First, we find the first derivative:
\( \frac{dy}{dx} = \frac{4x^3}{4} - \frac{2 \cdot 3x^2}{3} + \frac{2x}{2} + 0 \)
\( \frac{dy}{dx} = x^3 - 2x^2 + x \)
Next, we find the second derivative:
\( \frac{d^2 y}{d x^2} = 3x^2 - 4x + 1 \)
Now, we check the value of the first derivative at \( x = 1 \):
\( \left( \frac{dy}{dx} \right)_{x=1} = (1)^3 - 2(1)^2 + 1 = 1 - 2 + 1 = 0 \)
Since the first derivative is zero at \( x = 1 \), it is a critical point.

Next, we check the value of the second derivative at \( x = 1 \):
\( \left( \frac{d^2 y}{d x^2} \right)_{x=1} = 3(1)^2 - 4(1) + 1 = 3 - 4 + 1 = 0 \)
Since the second derivative is also zero at \( x = 1 \), it means we cannot use the second derivative test to determine if it's a maximum or minimum. We need to find the third derivative.
Let's find the third derivative:
\( \frac{d^3 y}{d x^3} = 6x - 4 \)
Now, we check the value of the third derivative at \( x = 1 \):
\( \left( \frac{d^3 y}{d x^3} \right)_{x=1} = 6(1) - 4 = 2 \)
Since the third derivative \( \left( \frac{d^3 y}{d x^3} \right)_{x=1} = 2 \neq 0 \), this indicates that \( x = 1 \) is a point of inflexion, not a local maximum or minimum.
At a point of inflexion, the curve changes its concavity. So, the ordinate at \( x=1 \) is neither a maximum nor a minimum.
In simple words: Even if the slope of a curve is flat at a point (meaning the first derivative is zero), it doesn't always mean it's a peak or a valley. If the second derivative is also zero, we check the third derivative. If the third derivative is not zero, the point is an "inflexion point" where the curve changes how it bends, rather than reaching a high or low point.

🎯 Exam Tip: Always remember that if the second derivative is zero at a critical point, you must use the third derivative test to determine if it's an inflexion point. A non-zero third derivative indicates an inflexion point.

Question 6. Find the points at which the function f given by \( f(x) = (x - 2)^4 (x + 1)^3 \) has
(i) local maxima
(ii) local minima
(iii) point of inflexion
Answer: Let the given function be \( f(x) = (x - 2)^4 (x + 1)^3 \).
First, we find the first derivative \( f'(x) \) using the product rule:
\( f'(x) = (x - 2)^4 \cdot 3(x + 1)^2 \cdot 1 + (x + 1)^3 \cdot 4(x - 2)^3 \cdot 1 \)
Factor out common terms, \( (x + 1)^2 (x - 2)^3 \):
\( f'(x) = (x + 1)^2 (x - 2)^3 [3(x - 2) + 4(x + 1)] \)
Simplify the expression inside the square brackets:
\( 3x - 6 + 4x + 4 = 7x - 2 \)
So, \( f'(x) = (x + 1)^2 (x - 2)^3 (7x - 2) \)
To find critical points, we set \( f'(x) = 0 \):
\( (x + 1)^2 (x - 2)^3 (7x - 2) = 0 \)
This gives us critical points at \( x = -1 \), \( x = 2 \), and \( x = \frac{2}{7} \).

Now, we use the first derivative test to determine the nature of these points.

**Case I: At \( x = -1 \)**
Consider points slightly less than -1, e.g., \( x = -1.1 \):
\( (x + 1)^2 \implies (-1.1 + 1)^2 = (-0.1)^2 = (+ve) \)
\( (x - 2)^3 \implies (-1.1 - 2)^3 = (-3.1)^3 = (-ve) \)
\( (7x - 2) \implies (7(-1.1) - 2) = (-7.7 - 2) = (-ve) \)
So, \( f'(x) = (+ve) \cdot (-ve) \cdot (-ve) = (+ve) \) when \( x < -1 \).
Consider points slightly greater than -1, e.g., \( x = -0.9 \):
\( (x + 1)^2 \implies (-0.9 + 1)^2 = (0.1)^2 = (+ve) \)
\( (x - 2)^3 \implies (-0.9 - 2)^3 = (-2.9)^3 = (-ve) \)
\( (7x - 2) \implies (7(-0.9) - 2) = (-6.3 - 2) = (-ve) \)
So, \( f'(x) = (+ve) \cdot (-ve) \cdot (-ve) = (+ve) \) when \( x > -1 \).
Since \( f'(x) \) does not change sign around \( x = -1 \), \( x = -1 \) is a point of inflexion. It is neither a local maximum nor a local minimum.

**Case II: At \( x = 2 \)**
Consider points slightly less than 2, e.g., \( x = 1.9 \):
\( (x + 1)^2 \implies (+ve) \)
\( (x - 2)^3 \implies (1.9 - 2)^3 = (-0.1)^3 = (-ve) \)
\( (7x - 2) \implies (7(1.9) - 2) = (13.3 - 2) = (+ve) \)
So, \( f'(x) = (+ve) \cdot (-ve) \cdot (+ve) = (-ve) \) when \( x < 2 \).
Consider points slightly greater than 2, e.g., \( x = 2.1 \):
\( (x + 1)^2 \implies (+ve) \)
\( (x - 2)^3 \implies (2.1 - 2)^3 = (0.1)^3 = (+ve) \)
\( (7x - 2) \implies (7(2.1) - 2) = (14.7 - 2) = (+ve) \)
So, \( f'(x) = (+ve) \cdot (+ve) \cdot (+ve) = (+ve) \) when \( x > 2 \).
Since \( f'(x) \) changes sign from negative to positive as x passes through 2, \( x = 2 \) is a point of local minima.

**Case III: At \( x = \frac{2}{7} \)**
Consider points slightly less than \( \frac{2}{7} \), e.g., \( x = 0.2 \) (since \( \frac{2}{7} \approx 0.28 \)):
\( (x + 1)^2 \implies (+ve) \)
\( (x - 2)^3 \implies (0.2 - 2)^3 = (-1.8)^3 = (-ve) \)
\( (7x - 2) \implies (7(0.2) - 2) = (1.4 - 2) = (-ve) \)
So, \( f'(x) = (+ve) \cdot (-ve) \cdot (-ve) = (+ve) \) when \( x < \frac{2}{7} \).
Consider points slightly greater than \( \frac{2}{7} \), e.g., \( x = 0.3 \):
\( (x + 1)^2 \implies (+ve) \)
\( (x - 2)^3 \implies (0.3 - 2)^3 = (-1.7)^3 = (-ve) \)
\( (7x - 2) \implies (7(0.3) - 2) = (2.1 - 2) = (+ve) \)
So, \( f'(x) = (+ve) \cdot (-ve) \cdot (+ve) = (-ve) \) when \( x > \frac{2}{7} \).
Since \( f'(x) \) changes sign from positive to negative as x passes through \( \frac{2}{7} \), \( x = \frac{2}{7} \) is a point of local maxima.

Summary:
(i) Local maxima: \( x = \frac{2}{7} \)
(ii) Local minima: \( x = 2 \)
(iii) Point of inflexion: \( x = -1 \)
In simple words: To find where a function has peaks, valleys, or changes its bend, we first find its derivative and set it to zero to get critical points. Then, we check the sign of the derivative just before and just after each critical point. If the sign changes from positive to negative, it's a peak (maxima). If it changes from negative to positive, it's a valley (minima). If the sign doesn't change, it's an inflexion point, where the curve changes how it bends.

🎯 Exam Tip: When using the first derivative test, clearly show the sign change of \( f'(x) \) around each critical point. Remember to state the conclusion for each type of point (maxima, minima, or inflexion).

Question 7. Discuss the maxima and minima of the expression \( \frac{6 x^3-45 x^2+108 x+2}{2 x^3-15 x^2+36 x+1} \).
Answer: Let the given expression be \( y \). We can simplify this expression using polynomial division or by noticing a relationship.
Let \( y = \frac{6 x^3-45 x^2+108 x+2}{2 x^3-15 x^2+36 x+1} \).
Notice that the numerator is \( 3 \times (2x^3 - 15x^2 + 36x + 1) - 1 \).
So, \( y = \frac{3(2x^3 - 15x^2 + 36x + 1) - 1}{2 x^3-15 x^2+36 x+1} \)
\( y = 3 - \frac{1}{2 x^3-15 x^2+36 x+1} \)
For y to have maxima or minima, the denominator \( u = 2x^3 - 15x^2 + 36x + 1 \) must have minima or maxima, respectively, because if \( u \) is maximum, \( \frac{1}{u} \) is minimum, making \( 3 - \frac{1}{u} \) maximum, and vice versa.

Let \( u(x) = 2x^3 - 15x^2 + 36x + 1 \).
First, find the derivative of \( u(x) \):
\( \frac{du}{dx} = 6x^2 - 30x + 36 \)
Next, find the second derivative of \( u(x) \):
\( \frac{d^2 u}{d x^2} = 12x - 30 \)
To find the critical points for \( u(x) \), set \( \frac{du}{dx} = 0 \):
\( \implies \) \( 6x^2 - 30x + 36 = 0 \)
Divide by 6: \( x^2 - 5x + 6 = 0 \)
Factor the quadratic equation: \( (x - 2)(x - 3) = 0 \)
This gives us two critical values for x: \( x = 2 \) or \( x = 3 \).

Now, we check the second derivative of \( u(x) \) at these points.
When \( x = 2 \):
\( \left( \frac{d^2 u}{d x^2} \right)_{x=2} = 12(2) - 30 = 24 - 30 = -6 \)
Since \( -6 < 0 \), \( x = 2 \) is a point of local maxima for \( u(x) \). This means \( u(x) \) is maximized at \( x=2 \).
When \( u(x) \) is maximum, \( \frac{1}{u(x)} \) is minimum, which makes \( y = 3 - \frac{1}{u(x)} \) maximum.
The value of \( u(x) \) at \( x=2 \) is: \( u(2) = 2(2^3) - 15(2^2) + 36(2) + 1 = 16 - 60 + 72 + 1 = 29 \).
The maximum value of \( y \) is \( 3 - \frac{1}{29} = \frac{87 - 1}{29} = \frac{86}{29} \).

When \( x = 3 \):
\( \left( \frac{d^2 u}{d x^2} \right)_{x=3} = 12(3) - 30 = 36 - 30 = 6 \)
Since \( 6 > 0 \), \( x = 3 \) is a point of local minima for \( u(x) \). This means \( u(x) \) is minimized at \( x=3 \).
When \( u(x) \) is minimum, \( \frac{1}{u(x)} \) is maximum, which makes \( y = 3 - \frac{1}{u(x)} \) minimum.
The value of \( u(x) \) at \( x=3 \) is: \( u(3) = 2(3^3) - 15(3^2) + 36(3) + 1 = 54 - 135 + 108 + 1 = 28 \).
The minimum value of \( y \) is \( 3 - \frac{1}{28} = \frac{84 - 1}{28} = \frac{83}{28} \).
In simple words: To find the highest and lowest points of a complex fraction, sometimes we can simplify it. Here, we rewrote the fraction so we just had to find the high and low points of its denominator. If the denominator is highest, the whole fraction is lowest, and vice versa. Then we used derivatives on the denominator to find its turning points and values.

🎯 Exam Tip: When dealing with rational functions, look for ways to simplify the expression first, often by polynomial division or algebraic manipulation, to make finding derivatives easier. Understanding how the numerator and denominator affect the overall function's maxima and minima is key.

Question 8. Find the turning values of the function \( -x^3 + 12x^2 - 5 \), distinguishing whether the value is a maximum, minimum or inflexional.
Answer: Let the given function be \( y = -x^3 + 12x^2 - 5 \).
First, we find the first derivative:
\( \frac{dy}{dx} = -3x^2 + 24x \)
Next, we find the second derivative:
\( \frac{d^2 y}{d x^2} = -6x + 24 \)
To find the critical points (turning points), we set the first derivative to zero:
\( \frac{dy}{dx} = 0 \)
\( \implies \) \( -3x^2 + 24x = 0 \)
Factor out \( -3x \): \( -3x(x - 8) = 0 \)
This gives us two critical values for x: \( x = 0 \) or \( x = 8 \).

Now, we check the second derivative at these points.
When \( x = 0 \):
\( \left( \frac{d^2 y}{d x^2} \right)_{x=0} = -6(0) + 24 = 24 \)
Since \( 24 > 0 \), \( x = 0 \) is a point of local minima.
The minimum value of y is:
\( y = -(0)^3 + 12(0)^2 - 5 = -5 \)
So, the minimum point is \( (0, -5) \).

When \( x = 8 \):
\( \left( \frac{d^2 y}{d x^2} \right)_{x=8} = -6(8) + 24 = -48 + 24 = -24 \)
Since \( -24 < 0 \), \( x = 8 \) is a point of local maxima.
The maximum value of y is:
\( y = -(8)^3 + 12(8)^2 - 5 \)
\( y = -512 + 12(64) - 5 \)
\( y = -512 + 768 - 5 \)
\( y = 251 \)
So, the maximum point is \( (8, 251) \).
In simple words: For any function, we can find its highest and lowest points by calculating its first and second derivatives. The first derivative tells us where the slope is flat, and the second derivative tells us if it's a peak (maximum) or a valley (minimum). We then plug these x-values back into the original function to find the actual high or low values.

🎯 Exam Tip: Be careful with signs when differentiating negative terms. Always show both the x and y coordinates of the turning points, which are found by substituting the critical x-values back into the original function.