Mechanical Engineering Exam > Mechanical Engineering Questions > The function f(x) = x3- 6x2+ 9x+25 hasa)A max...
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The function f(x) = x3- 6x2+ 9x+25 has
- a)A maxima at x = 1 and a minima at x = 3
- b)A maxima at x = 3 and a minima at x = 1
- c)No maxima, but a minima at x = 3
- d)A maxima at x = 1, but not minima
Correct answer is option 'A'. Can you explain this answer?
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The function f(x) = x3- 6x2+ 9x+25 hasa)A maxima at x = 1 and a minima...
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The function f(x) = x3- 6x2+ 9x+25 hasa)A maxima at x = 1 and a minima...
Explanation:
To determine the maxima and minima of a function, we need to find the critical points. Critical points occur where the derivative of the function is equal to zero or does not exist.
Finding the derivative:
Let's find the derivative of the given function f(x) = x^3 - 6x^2 + 9x + 25:
f'(x) = 3x^2 - 12x + 9
Finding the critical points:
To find the critical points, we need to solve the equation f'(x) = 0:
3x^2 - 12x + 9 = 0
Simplifying the equation, we get:
x^2 - 4x + 3 = 0
Factoring the quadratic equation, we have:
(x - 1)(x - 3) = 0
So the critical points are x = 1 and x = 3.
Determining the nature of the critical points:
To determine whether the critical points are maxima or minima, we can use the second derivative test. If the second derivative is positive at a critical point, then it is a minima. If the second derivative is negative, then it is a maxima.
Finding the second derivative:
Let's find the second derivative of f(x):
f''(x) = d/dx(3x^2 - 12x + 9)
= 6x - 12
Evaluating the second derivative at the critical points:
Substituting x = 1 and x = 3 into f''(x), we get:
f''(1) = 6(1) - 12 = -6
f''(3) = 6(3) - 12 = 6
Conclusion:
From the second derivative test, we can conclude the nature of the critical points as follows:
- At x = 1, f''(1) = -6, which means the function has a maximum at x = 1.
- At x = 3, f''(3) = 6, which means the function has a minimum at x = 3.
Therefore, the correct answer is option A: The function has a maxima at x = 1 and a minima at x = 3.
To determine the maxima and minima of a function, we need to find the critical points. Critical points occur where the derivative of the function is equal to zero or does not exist.
Finding the derivative:
Let's find the derivative of the given function f(x) = x^3 - 6x^2 + 9x + 25:
f'(x) = 3x^2 - 12x + 9
Finding the critical points:
To find the critical points, we need to solve the equation f'(x) = 0:
3x^2 - 12x + 9 = 0
Simplifying the equation, we get:
x^2 - 4x + 3 = 0
Factoring the quadratic equation, we have:
(x - 1)(x - 3) = 0
So the critical points are x = 1 and x = 3.
Determining the nature of the critical points:
To determine whether the critical points are maxima or minima, we can use the second derivative test. If the second derivative is positive at a critical point, then it is a minima. If the second derivative is negative, then it is a maxima.
Finding the second derivative:
Let's find the second derivative of f(x):
f''(x) = d/dx(3x^2 - 12x + 9)
= 6x - 12
Evaluating the second derivative at the critical points:
Substituting x = 1 and x = 3 into f''(x), we get:
f''(1) = 6(1) - 12 = -6
f''(3) = 6(3) - 12 = 6
Conclusion:
From the second derivative test, we can conclude the nature of the critical points as follows:
- At x = 1, f''(1) = -6, which means the function has a maximum at x = 1.
- At x = 3, f''(3) = 6, which means the function has a minimum at x = 3.
Therefore, the correct answer is option A: The function has a maxima at x = 1 and a minima at x = 3.
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The function f(x) = x3- 6x2+ 9x+25 hasa)A maxima at x = 1 and a minima at x = 3 b)A maxima at x = 3 and a minima at x = 1 c)No maxima, but a minima at x = 3 d)A maxima at x = 1, but not minimaCorrect answer is option 'A'. Can you explain this answer?
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