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Consider the function f(x) = x2 – x – 2. The maximum value of f(x) in the closed interval [–4, 4] is
- a)18
- b)10
- c)–2.25
- d)indete rminate
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Consider the function f(x) = x2 – x – 2. The maximum value...
∴ f (x )has minimum at x= 1 / 2 It Shows that a maximum value that will be at x = 4 or x = - 4
At x = 4, f (x )= 10
∴ At x= −4, f (x ) = 18
∴ At x= −4, f (x ) has a maximum.
Most Upvoted Answer
Consider the function f(x) = x2 – x – 2. The maximum value...
Introduction:
We are given a function f(x) = x^2 - x + 2 and we need to find the maximum value of f(x) in the closed interval [4, 4]. To find the maximum value, we can use the first and second derivative tests.
First Derivative Test:
To find the critical points of f(x), we need to find the values of x where the derivative of f(x) is equal to zero or undefined.
The derivative of f(x) is given by f'(x) = 2x - 1.
Setting f'(x) = 0, we get:
2x - 1 = 0
2x = 1
x = 1/2
Since x = 1/2 is not in the closed interval [4, 4], it is not a critical point.
Second Derivative Test:
To determine the nature of the critical point x = 1/2, we need to find the second derivative of f(x).
The second derivative of f(x) is given by f''(x) = 2.
Since f''(x) = 2 > 0, the critical point x = 1/2 is a local minimum.
Endpoints of the Interval:
Next, we need to evaluate the function at the endpoints of the interval [4, 4].
f(4) = 4^2 - 4 + 2 = 16 - 4 + 2 = 14
f(4) = 14
Conclusion:
Since the critical point x = 1/2 is a local minimum and the value of f(x) at the endpoints is 14, the maximum value of f(x) in the closed interval [4, 4] is 14.
Therefore, the correct answer is option A) 18.
We are given a function f(x) = x^2 - x + 2 and we need to find the maximum value of f(x) in the closed interval [4, 4]. To find the maximum value, we can use the first and second derivative tests.
First Derivative Test:
To find the critical points of f(x), we need to find the values of x where the derivative of f(x) is equal to zero or undefined.
The derivative of f(x) is given by f'(x) = 2x - 1.
Setting f'(x) = 0, we get:
2x - 1 = 0
2x = 1
x = 1/2
Since x = 1/2 is not in the closed interval [4, 4], it is not a critical point.
Second Derivative Test:
To determine the nature of the critical point x = 1/2, we need to find the second derivative of f(x).
The second derivative of f(x) is given by f''(x) = 2.
Since f''(x) = 2 > 0, the critical point x = 1/2 is a local minimum.
Endpoints of the Interval:
Next, we need to evaluate the function at the endpoints of the interval [4, 4].
f(4) = 4^2 - 4 + 2 = 16 - 4 + 2 = 14
f(4) = 14
Conclusion:
Since the critical point x = 1/2 is a local minimum and the value of f(x) at the endpoints is 14, the maximum value of f(x) in the closed interval [4, 4] is 14.
Therefore, the correct answer is option A) 18.
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Consider the function f(x) = x2 – x – 2. The maximum value of f(x) in the closed interval [–4, 4] isa)18 b)10 c)–2.25 d)indete rminateCorrect answer is option 'A'. Can you explain this answer?
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