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The greatest and the least value of the function f(x) = x3 – 18x2 + 96x in the interval [0, 9] are :
Correct answer is '160'. Can you explain this answer?
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Verified Answer
The greatest and the least value of the functionf(x) =x3– 18x2+ ...
f(x) = x3 – 18x2 + 96x
= 3(x – 4)(x – 8)
Put f'(x) = 0, we get x = 4, 8
f''(x) = 6x – 36 = 6(x – 6)
which is positive at x = 8 and negative at x = 4
= 3(x – 4)(x – 8)
Put f'(x) = 0, we get x = 4, 8
f''(x) = 6x – 36 = 6(x – 6)
which is positive at x = 8 and negative at x = 4
Now, f(0) = 0, f(8) = 128
f(4) = 160, f(9) = 135
∴ least value is 0 and greatest value is 160.
f(4) = 160, f(9) = 135
∴ least value is 0 and greatest value is 160.
The correct answer is: 160
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The greatest and the least value of the functionf(x) =x3– 18x2+ ...
Finding the Values of the Function
To analyze the function f(x) = x^3 - 18x^2 + 96x within the interval [0, 9], we need to determine its critical points and evaluate the function at both the endpoints of the interval.
Step 1: Find the Derivative
- The first step is to compute the derivative of the function:
f'(x) = 3x^2 - 36x + 96
Step 2: Set the Derivative to Zero
- To find critical points, we set the derivative equal to zero:
3x^2 - 36x + 96 = 0
- Dividing the equation by 3 simplifies it to:
x^2 - 12x + 32 = 0
- This can be factored to:
(x - 4)(x - 8) = 0
- Thus, the critical points are x = 4 and x = 8.
Step 3: Evaluate the Function
- Next, we need to evaluate the function at the endpoints and critical points:
- f(0) = 0^3 - 18(0)^2 + 96(0) = 0
- f(4) = 4^3 - 18(4)^2 + 96(4) = 64 - 288 + 384 = 160
- f(8) = 8^3 - 18(8)^2 + 96(8) = 512 - 1152 + 768 = 128
Step 4: Determine the Greatest and Least Values
- Comparing the values:
- f(0) = 0
- f(4) = 160 (greatest value)
- f(8) = 128
- Therefore, the least value is 0, and the greatest value of the function in the interval [0, 9] is indeed 160.
To analyze the function f(x) = x^3 - 18x^2 + 96x within the interval [0, 9], we need to determine its critical points and evaluate the function at both the endpoints of the interval.
Step 1: Find the Derivative
- The first step is to compute the derivative of the function:
f'(x) = 3x^2 - 36x + 96
Step 2: Set the Derivative to Zero
- To find critical points, we set the derivative equal to zero:
3x^2 - 36x + 96 = 0
- Dividing the equation by 3 simplifies it to:
x^2 - 12x + 32 = 0
- This can be factored to:
(x - 4)(x - 8) = 0
- Thus, the critical points are x = 4 and x = 8.
Step 3: Evaluate the Function
- Next, we need to evaluate the function at the endpoints and critical points:
- f(0) = 0^3 - 18(0)^2 + 96(0) = 0
- f(4) = 4^3 - 18(4)^2 + 96(4) = 64 - 288 + 384 = 160
- f(8) = 8^3 - 18(8)^2 + 96(8) = 512 - 1152 + 768 = 128
Step 4: Determine the Greatest and Least Values
- Comparing the values:
- f(0) = 0
- f(4) = 160 (greatest value)
- f(8) = 128
- Therefore, the least value is 0, and the greatest value of the function in the interval [0, 9] is indeed 160.
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The greatest and the least value of the functionf(x) =x3– 18x2+ 96xin the interval [0, 9] are :Correct answer is '160'. Can you explain this answer?
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The greatest and the least value of the functionf(x) =x3– 18x2+ 96xin the interval [0, 9] are :Correct answer is '160'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about The greatest and the least value of the functionf(x) =x3– 18x2+ 96xin the interval [0, 9] are :Correct answer is '160'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The greatest and the least value of the functionf(x) =x3– 18x2+ 96xin the interval [0, 9] are :Correct answer is '160'. Can you explain this answer?.
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