Question:
The function $f(x)=e^x-x$ is increasing in the interval:
The function $f(x)=e^x-x$ is increasing in the interval:
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To find decreasing intervals, solve for $f'(x) < 0$.
Updated On: Apr 28, 2026
- (0, 4)
- $(-\infty, 0)$
- (-1, 1)
- (-1, 0)
- $(0, \infty)$
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The Correct Option is
Solution and Explanation
Step 1: Concept
A function is increasing where its derivative $f'(x) > 0$.
Step 2: Analysis
$f(x) = e^x - x \implies f'(x) = e^x - 1$. Set $e^x - 1 > 0 \implies e^x > 1$.
Step 3: Conclusion
Since $1 = e^0$, the condition $e^x > e^0$ implies $x > 0$. Therefore, the function is increasing in the interval $(0, \infty)$. Final Answer: (E)
A function is increasing where its derivative $f'(x) > 0$.
Step 2: Analysis
$f(x) = e^x - x \implies f'(x) = e^x - 1$. Set $e^x - 1 > 0 \implies e^x > 1$.
Step 3: Conclusion
Since $1 = e^0$, the condition $e^x > e^0$ implies $x > 0$. Therefore, the function is increasing in the interval $(0, \infty)$. Final Answer: (E)
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