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The maximum and the minimum value of 3x4 – 8x3 + 12x2 – 48x + 1 on the interval [1,4]
- a)-40,257
- b)-48,258
- c)-49,258
- d)-58,257
Correct answer is option 'A'. Can you explain this answer?
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The maximum and the minimum value of 3x4– 8x3+ 12x2– 48x +...
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The maximum and the minimum value of 3x4– 8x3+ 12x2– 48x +...
Solution:
To find the maximum and minimum value of the given function, we need to differentiate it first.
f(x) = 3x^4 + 8x^3 + 12x^2 + 48x + 1
f'(x) = 12x^3 + 24x^2 + 24x + 48
Now, we need to find the critical points of the function by equating f'(x) to zero.
12x^3 + 24x^2 + 24x + 48 = 0
Dividing both sides by 12, we get
x^3 + 2x^2 + 2x + 4 = 0
We can see that x = -2 is a root of the above equation.
Using synthetic division, we get
x^2 + 4x + 2 = 0
Using the quadratic formula, we get
x = (-4 ± sqrt(16 - 8)) / 2
x = -2 ± sqrt(2)
We have two critical points:
x = -2 - sqrt(2) and x = -2 + sqrt(2)
Now, we need to find the value of the function at these critical points and at the endpoints of the interval [1,4].
f(1) = 3 + 8 + 12 + 48 + 1 = 72
f(4) = 192 + 256 + 192 + 192 + 1 = 833
f(-2 - sqrt(2)) ≈ -40.28
f(-2 + sqrt(2)) ≈ -49.26
Therefore, the maximum value of the function on the interval [1,4] is 833 and the minimum value is -40.28, which is closest to option A.
To find the maximum and minimum value of the given function, we need to differentiate it first.
f(x) = 3x^4 + 8x^3 + 12x^2 + 48x + 1
f'(x) = 12x^3 + 24x^2 + 24x + 48
Now, we need to find the critical points of the function by equating f'(x) to zero.
12x^3 + 24x^2 + 24x + 48 = 0
Dividing both sides by 12, we get
x^3 + 2x^2 + 2x + 4 = 0
We can see that x = -2 is a root of the above equation.
Using synthetic division, we get
x^2 + 4x + 2 = 0
Using the quadratic formula, we get
x = (-4 ± sqrt(16 - 8)) / 2
x = -2 ± sqrt(2)
We have two critical points:
x = -2 - sqrt(2) and x = -2 + sqrt(2)
Now, we need to find the value of the function at these critical points and at the endpoints of the interval [1,4].
f(1) = 3 + 8 + 12 + 48 + 1 = 72
f(4) = 192 + 256 + 192 + 192 + 1 = 833
f(-2 - sqrt(2)) ≈ -40.28
f(-2 + sqrt(2)) ≈ -49.26
Therefore, the maximum value of the function on the interval [1,4] is 833 and the minimum value is -40.28, which is closest to option A.
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The maximum and the minimum value of 3x4– 8x3+ 12x2– 48x + 1 on the interval [1,4]a)-40,257b)-48,258c)-49,258d)-58,257Correct answer is option 'A'. Can you explain this answer?
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