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The function f(x) = x3 - 3x is ....... .
- a)increasing in ( -∞, - 1) ∪ ( 1, ∞) and decreasing in (-1, 1)
- b)increasing in (0, ∞) and decreasing in (-∞, 0)
- c)decreasing in(0, ∞) and increasing in (-∞, 0)
- d)decreasing in (-∞, -1) ∪ (1, ∞)and increasing in (-1, 1)
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The function f(x) = x3 - 3x is ....... .a)increasing in ( -∞, - 1) ∪ ...
We have, function f(x) = x3 - 3x
⇒f’(x) = 3x2 - 3 = 3(x + 1)(x - 1)
f’(x) = 0 ⇒ x = -1,1
Hence, f'(x) > 0 in (-∞, -1) ∪ (1, ∞) and f'(x) < o="" in="" (-1,="" />
∴ f(x) is increasing in (-∞, -1), ∪(1, ∞) and decreasing in (-1, 1).
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
The function f(x) = x3 - 3x is ....... .a)increasing in ( -∞, - 1) ∪ ...
Explanation:
Increasing and Decreasing Intervals:
- To determine where the function is increasing or decreasing, we need to find the critical points of the function.
- The critical points occur where the derivative of the function is equal to zero or undefined.
- Taking the derivative of f(x) = x^3 - 3x, we get f'(x) = 3x^2 - 3.
- Setting f'(x) = 0, we find the critical points at x = -1 and x = 1.
- Now, we can test the intervals between these critical points to determine where the function is increasing or decreasing.
Interval Analysis:
- For x < -1,="" choose="" x="-2." substituting="" x="-2" into="" f'(x),="" we="" get="" f'(-2)="3(-2)^2" -="" 3="9" -="" 3="6." this="" means="" f(x)="" is="" increasing="" in="" (-∞,="" />
- For -1 < x="" />< 1,="" choose="" x="0." substituting="" x="0" into="" f'(x),="" we="" get="" f'(0)="3(0)^2" -="" 3="-3." this="" means="" f(x)="" is="" decreasing="" in="" (-1,="" />
- For x > 1, choose x = 2. Substituting x = 2 into f'(x), we get f'(2) = 3(2)^2 - 3 = 9 - 3 = 6. This means f(x) is increasing in (1, ∞).
Therefore, the function f(x) = x^3 - 3x is increasing in (-∞, -1) ∪ (1, ∞) and decreasing in (-1, 1), which matches with option 'A'.
Increasing and Decreasing Intervals:
- To determine where the function is increasing or decreasing, we need to find the critical points of the function.
- The critical points occur where the derivative of the function is equal to zero or undefined.
- Taking the derivative of f(x) = x^3 - 3x, we get f'(x) = 3x^2 - 3.
- Setting f'(x) = 0, we find the critical points at x = -1 and x = 1.
- Now, we can test the intervals between these critical points to determine where the function is increasing or decreasing.
Interval Analysis:
- For x < -1,="" choose="" x="-2." substituting="" x="-2" into="" f'(x),="" we="" get="" f'(-2)="3(-2)^2" -="" 3="9" -="" 3="6." this="" means="" f(x)="" is="" increasing="" in="" (-∞,="" />
- For -1 < x="" />< 1,="" choose="" x="0." substituting="" x="0" into="" f'(x),="" we="" get="" f'(0)="3(0)^2" -="" 3="-3." this="" means="" f(x)="" is="" decreasing="" in="" (-1,="" />
- For x > 1, choose x = 2. Substituting x = 2 into f'(x), we get f'(2) = 3(2)^2 - 3 = 9 - 3 = 6. This means f(x) is increasing in (1, ∞).
Therefore, the function f(x) = x^3 - 3x is increasing in (-∞, -1) ∪ (1, ∞) and decreasing in (-1, 1), which matches with option 'A'.
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The function f(x) = x3 - 3x is ....... .a)increasing in ( -∞, - 1) ∪ ( 1, ∞) and decreasing in (-1, 1)b)increasing in (0, ∞) and decreasing in (-∞, 0)c)decreasing in(0, ∞) and increasing in (-∞, 0)d)decreasing in (-∞, -1) ∪ (1, ∞)and increasing in (-1, 1)Correct answer is option 'A'. Can you explain this answer?
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