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The function f(x) = 2x3 – 3x2 – 12x + 4, has
- a)two points of local maximum
- b)two points of local minimum
- c)one maxima and one minima
- d)no maxima or minima
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The function f(x) = 2x3 – 3x2 – 12x + 4, hasa)two points o...
f(x) = 2x3 – 3x2 – 12x + 4
⇒ f '(x) = 6x2 – 6x – 12 = 6(x2 – x – 2)
= 6(x – 2) (x + 1)
For maxima and minima f '(x) = 0
∴ 6(x – 2)(x + 1) = 0 Þ x = 2, – 1
Now, f ''(x) = 12x-6
At x = 2; f ''(x) = 24 - 6 = 18> 0
∴ x = 2 , local min. point
At x = – 1; f ''(x) = 12 ( -1) - 6 = -18<0
∴ x = –1 local max. point
⇒ f '(x) = 6x2 – 6x – 12 = 6(x2 – x – 2)
= 6(x – 2) (x + 1)
For maxima and minima f '(x) = 0
∴ 6(x – 2)(x + 1) = 0 Þ x = 2, – 1
Now, f ''(x) = 12x-6
At x = 2; f ''(x) = 24 - 6 = 18> 0
∴ x = 2 , local min. point
At x = – 1; f ''(x) = 12 ( -1) - 6 = -18<0
∴ x = –1 local max. point
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The function f(x) = 2x3 – 3x2 – 12x + 4, hasa)two points o...
Understanding the Function
The function given is f(x) = 2x^3 – 3x^2 – 12x + 4. To analyze its local maxima and minima, we first need to find the critical points by calculating the first derivative.
Finding the First Derivative
- The first derivative f'(x) = 6x^2 - 6x - 12.
- Setting f'(x) = 0 gives us the critical points.
Finding Critical Points
- We solve the equation: 6x^2 - 6x - 12 = 0.
- Simplifying, we get: x^2 - x - 2 = 0.
- Factoring provides the solutions: (x - 2)(x + 1) = 0, yielding x = 2 and x = -1.
Determining Local Extrema
- To classify the critical points, we use the second derivative test.
- The second derivative f''(x) = 12x - 6.
Evaluating the Second Derivative
- For x = 2: f''(2) = 12(2) - 6 = 24, which is positive. This indicates a local minimum.
- For x = -1: f''(-1) = 12(-1) - 6 = -18, which is negative. This indicates a local maximum.
Conclusion
- The function has one local maximum at x = -1 and one local minimum at x = 2.
- Hence, the correct answer is option 'C': one maxima and one minima.
This analysis shows that f(x) exhibits one peak and one trough, confirming the classification of its critical points effectively.
The function given is f(x) = 2x^3 – 3x^2 – 12x + 4. To analyze its local maxima and minima, we first need to find the critical points by calculating the first derivative.
Finding the First Derivative
- The first derivative f'(x) = 6x^2 - 6x - 12.
- Setting f'(x) = 0 gives us the critical points.
Finding Critical Points
- We solve the equation: 6x^2 - 6x - 12 = 0.
- Simplifying, we get: x^2 - x - 2 = 0.
- Factoring provides the solutions: (x - 2)(x + 1) = 0, yielding x = 2 and x = -1.
Determining Local Extrema
- To classify the critical points, we use the second derivative test.
- The second derivative f''(x) = 12x - 6.
Evaluating the Second Derivative
- For x = 2: f''(2) = 12(2) - 6 = 24, which is positive. This indicates a local minimum.
- For x = -1: f''(-1) = 12(-1) - 6 = -18, which is negative. This indicates a local maximum.
Conclusion
- The function has one local maximum at x = -1 and one local minimum at x = 2.
- Hence, the correct answer is option 'C': one maxima and one minima.
This analysis shows that f(x) exhibits one peak and one trough, confirming the classification of its critical points effectively.
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The function f(x) = 2x3 – 3x2 – 12x + 4, hasa)two points of local maximumb)two points of local minimumc)one maxima and one minimad)no maxima or minimaCorrect answer is option 'C'. Can you explain this answer?
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