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The minimum value of the function f(x) = 1/3x(x2 - 3) in the interval - 100 ≤ x ≤ 100 occurs at x = ______.
(Answer up to the nearest integer)
Correct answer is '-100'. Can you explain this answer?
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The minimum value of the function f(x) = 1/3x(x2 - 3) in the interval ...
To find the minimum value of the function f(x) = 1/3x(x^2 - 3) in the interval -100 < x="" />< 100,="" we="" can="" start="" by="" finding="" the="" critical="" points="" of="" the="" />
To find the critical points, we need to find the values of x where the derivative of the function is equal to zero.
f'(x) = (1/3)(x^2 - 3) + (1/3)x(2x) = (1/3)(x^2 - 3 + 2x^2) = (1/3)(3x^2 - 3) = x^2 - 1
Setting f'(x) = 0, we have:
x^2 - 1 = 0
x^2 = 1
x = ±1
So the critical points are x = -1 and x = 1.
Next, we need to evaluate the function at these critical points as well as the endpoints of the interval.
f(-100) = (1/3)(-100)(-100^2 - 3) = (1/3)(-100)(1000000 - 3) = (1/3)(-100)(999997) = -33333300
f(-1) = (1/3)(-1)(-1^2 - 3) = (1/3)(-1)(-1 - 3) = (1/3)(-1)(-4) = 4/3
f(1) = (1/3)(1)(1^2 - 3) = (1/3)(1)(1 - 3) = (1/3)(1)(-2) = -2/3
f(100) = (1/3)(100)(100^2 - 3) = (1/3)(100)(1000000 - 3) = (1/3)(100)(999997) = 33333300
Comparing these values, we can see that the minimum value of the function occurs at x = -1, where f(-1) = 4/3.
Therefore, the minimum value of the function f(x) = 1/3x(x^2 - 3) in the interval -100 < x="" />< 100="" is="" 4/3.="" 100="" is="" />
To find the critical points, we need to find the values of x where the derivative of the function is equal to zero.
f'(x) = (1/3)(x^2 - 3) + (1/3)x(2x) = (1/3)(x^2 - 3 + 2x^2) = (1/3)(3x^2 - 3) = x^2 - 1
Setting f'(x) = 0, we have:
x^2 - 1 = 0
x^2 = 1
x = ±1
So the critical points are x = -1 and x = 1.
Next, we need to evaluate the function at these critical points as well as the endpoints of the interval.
f(-100) = (1/3)(-100)(-100^2 - 3) = (1/3)(-100)(1000000 - 3) = (1/3)(-100)(999997) = -33333300
f(-1) = (1/3)(-1)(-1^2 - 3) = (1/3)(-1)(-1 - 3) = (1/3)(-1)(-4) = 4/3
f(1) = (1/3)(1)(1^2 - 3) = (1/3)(1)(1 - 3) = (1/3)(1)(-2) = -2/3
f(100) = (1/3)(100)(100^2 - 3) = (1/3)(100)(1000000 - 3) = (1/3)(100)(999997) = 33333300
Comparing these values, we can see that the minimum value of the function occurs at x = -1, where f(-1) = 4/3.
Therefore, the minimum value of the function f(x) = 1/3x(x^2 - 3) in the interval -100 < x="" />< 100="" is="" 4/3.="" 100="" is="" />
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Community Answer
The minimum value of the function f(x) = 1/3x(x2 - 3) in the interval ...
⇒ x
2
- 1 = 0
⇒ x = ±1
f''(x) = 2x
f''(1) = 2 > 0 ⇒ at x = 1, f(x) has local minimum.
f''(- 1) = - 2 < 0 ⇒ at x = - 1, f(x) has local maximum
For x = 1 local minimum value = f(1) = 1/3 - 1 = -2/3
Finding f(- 100) = - 333433.33
f(100) = 333233.33
( ∵ x = 100, - 100 are end points of interval)
∴ Minimum occurs at x = - 100
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The minimum value of the function f(x) = 1/3x(x2 - 3) in the interval - 100 ≤ x ≤ 100 occurs at x = ______.(Answer up to the nearest integer)Correct answer is '-100'. Can you explain this answer?
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