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If f(x) = ln(x) + x^2, then the function is increasing for:
If f(x) = ln(x) + x^2, then the function is increasing for:
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Practice Questions
1 question
Q1
If f(x) = ln(x) + x^2, then the function is increasing for:
x > 0
x < 0
x > 1
x < 1
Show Solution
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The derivative f'(x) = 1/x + 2x. For f'(x) > 0, we need 1/x + 2x > 0, which holds for x > 0.
Questions & Step-by-step Solutions
1 item
Q
Q: If f(x) = ln(x) + x^2, then the function is increasing for:
Solution:
The derivative f'(x) = 1/x + 2x. For f'(x) > 0, we need 1/x + 2x > 0, which holds for x > 0.
Steps: 7
Show Steps
Step 1: Identify the function f(x) = ln(x) + x^2.
Step 2: Find the derivative of the function, which is f'(x).
Step 3: The derivative f'(x) is calculated as f'(x) = 1/x + 2x.
Step 4: To determine when the function is increasing, we need to find when f'(x) > 0.
Step 5: Set up the inequality: 1/x + 2x > 0.
Step 6: Analyze the inequality. Since 1/x is positive for x > 0 and 2x is also positive for x > 0, the entire expression is positive for x > 0.
Step 7: Conclude that the function f(x) is increasing for x > 0.
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