For the function $ f(x) = 2x^3 - 9x^2 + 12x - 5 $, $ x \in [0, 3] $, match List-I with List-II:List-IList-II(A) Absolute maximum value(I) 3(B) Absolute minimum value(II) 0(C) Point of maxima(III) -5(D) Point of minima(IV) 4Choose the correct answer from the options given below:

Question

Prepp Team · Accepted Answer

Finding Absolute Maximum and Minimum Values

To find the absolute maximum and minimum values of a function $ f(x) $ on a closed interval $ [a, b] $, we need to evaluate the function at the critical points within the interval and at the endpoints of the interval. The largest value among these is the absolute maximum value, and the smallest value is the absolute minimum value.

The given function is $ f(x) = 2x^3 - 9x^2 + 12x - 5 $, and the interval is $ [0, 3] $.

Step 1: Find the derivative of the function

The derivative of $ f(x) $ with respect to $ x $ is $ f'(x) $:

$ f'(x) = \frac{d}{dx} (2x^3 - 9x^2 + 12x - 5) $

$ f'(x) = 6x^2 - 18x + 12 $

Step 2: Find the critical points

Critical points are the points where $ f'(x) = 0 $ or $ f'(x) $ is undefined. In this case, $ f'(x) $ is a polynomial, so it is always defined. We set $ f'(x) = 0 $:

$ 6x^2 - 18x + 12 = 0 $

Divide the equation by 6:

$ x^2 - 3x + 2 = 0 $

Factor the quadratic equation:

$ (x - 1)(x - 2) = 0 $

The critical points are $ x = 1 $ and $ x = 2 $.

Step 3: Check if critical points are within the interval

The given interval is $ [0, 3] $.

Is $ x = 1 $ in $ [0, 3] $? Yes.
Is $ x = 2 $ in $ [0, 3] $? Yes.

Both critical points are within the interval.

Step 4: Evaluate the function at critical points and endpoints

We need to evaluate $ f(x) $ at the critical points $ x = 1, x = 2 $ and the endpoints $ x = 0, x = 3 $.

At $ x = 0 $ (Endpoint):
$ f(0) = 2(0)^3 - 9(0)^2 + 12(0) - 5 = 0 - 0 + 0 - 5 = -5 $
At $ x = 1 $ (Critical Point):
$ f(1) = 2(1)^3 - 9(1)^2 + 12(1) - 5 = 2 - 9 + 12 - 5 = 0 $
At $ x = 2 $ (Critical Point):
$ f(2) = 2(2)^3 - 9(2)^2 + 12(2) - 5 = 2(8) - 9(4) + 24 - 5 = 16 - 36 + 24 - 5 = -1 $
At $ x = 3 $ (Endpoint):
$ f(3) = 2(3)^3 - 9(3)^2 + 12(3) - 5 = 2(27) - 9(81) + 36 - 5 = 54 - 81 + 36 - 5 = 4 $

Step 5: Determine Absolute Maximum and Minimum Values

The values of the function at the relevant points are $ f(0) = -5 $, $ f(1) = 0 $, $ f(2) = -1 $, and $ f(3) = 4 $.

Comparing these values: $ \{ -5, 0, -1, 4 \} $

The largest value is 4. This is the absolute maximum value.
The smallest value is -5. This is the absolute minimum value.

Step 6: Determine Points of Maxima and Minima

The point of maxima is the x-value where the absolute maximum value occurs.

The absolute maximum value is 4, which occurs at $ x = 3 $. So, the point of maxima is 3.

The point of minima is the x-value where the absolute minimum value occurs.

The absolute minimum value is -5, which occurs at $ x = 0 $. So, the point of minima is 0.

Step 7: Match List-I with List-II

Based on our findings:

(A) Absolute maximum value is 4.
(B) Absolute minimum value is -5.
(C) Point of maxima is 3.
(D) Point of minima is 0.

Matching with List-II:

(A) Absolute maximum value (4) matches with (IV) 4.
(B) Absolute minimum value (-5) matches with (III) -5.
(C) Point of maxima (3) matches with (I) 3.
(D) Point of minima (0) matches with (II) 0.

So, the correct matching is (A) - (IV), (B) - (III), (C) - (I), (D) - (II).

Item (List-I)	Calculated Value/Point	Match (List-II)
(A) Absolute maximum value	4	(IV) 4
(B) Absolute minimum value	-5	(III) -5
(C) Point of maxima	3	(I) 3
(D) Point of minima	0	(II) 0

The correct matching is (A) - (IV), (B) - (III), (C) - (I), (D) - (II).

Revision Table for Function Extrema

Concept	Definition/How to Find	Relevance to Problem
Absolute Maximum Value	The highest value a function attains over a given interval. Found by evaluating the function at critical points within the interval and endpoints, and taking the maximum of these values.	We found the absolute maximum value of $f(x)$ on $ [0, 3] $ to be 4.
Absolute Minimum Value	The lowest value a function attains over a given interval. Found by evaluating the function at critical points within the interval and endpoints, and taking the minimum of these values.	We found the absolute minimum value of $f(x)$ on $ [0, 3] $ to be -5.
Critical Point	A point in the domain of the function where the derivative is either zero or undefined. These are potential locations for local maxima or minima.	We found critical points at $ x=1 $ and $ x=2 $ by setting $ f'(x) = 0 $.
Point of Maxima/Minima	The x-value (or input) where the absolute maximum or minimum value occurs.	The point of maxima is $ x=3 $ (where $f(x)=4$). The point of minima is $ x=0 $ (where $f(x)=-5$).
Closed Interval Method	The method used to find absolute extrema on a closed interval $ [a, b] $. It involves finding critical points in $ (a, b) $ and evaluating the function at these points and the endpoints $ a $ and $ b $.	This problem requires applying the closed interval method.

Additional Information on Function Extrema

Understanding extrema is a fundamental concept in calculus with many applications in various fields like physics, engineering, and economics. Extrema (plural of extremum) refer to the maximum and minimum values of a function.

Local vs. Absolute Extrema

Local Extrema: These are the maximum or minimum values of the function within a specific neighborhood or open interval. A function can have multiple local maxima and minima. Local extrema occur at critical points.
Absolute Extrema: Also known as global extrema, these are the overall maximum and minimum values of the function over its entire domain or a specified interval. A function on a closed and bounded interval is guaranteed to have both an absolute maximum and an absolute minimum value (Extreme Value Theorem).

Extreme Value Theorem

The Extreme Value Theorem states that if a function $ f $ is continuous on a closed interval $ [a, b] $, then $ f $ must attain both an absolute maximum value and an absolute minimum value on $ [a, b] $. Our function $ f(x) = 2x^3 - 9x^2 + 12x - 5 $ is a polynomial, which is continuous everywhere, including the closed interval $ [0, 3] $. Therefore, we are guaranteed to find absolute maximum and minimum values on this interval.

Finding Extrema Steps Summarized

Check if the function is continuous on the closed interval $ [a, b] $. (If yes, absolute extrema exist).
Find the derivative $ f'(x) $.
Find the critical points by solving $ f'(x) = 0 $ or finding where $ f'(x) $ is undefined.
Identify the critical points that lie within the open interval $ (a, b) $.
Evaluate the function $ f(x) $ at the critical points found in step 4 and at the endpoints $ a $ and $ b $.
The largest of these values is the absolute maximum value, and the smallest is the absolute minimum value.

The points of maxima and minima are the x-values where these absolute maximum and minimum values occur, respectively.

Prepp Team · Answer

(A) - (IV), (B) - (II), (C) - (I), (D) - (III)

Prepp Team · Answer

(A) - (II), (B) - (III), (C) - (I), (D) - (IV)

Prepp Team · Answer

(A) - (IV), (B) - (II), (C) - (III), (D) - (I)