Question

The function $f(x)=e^x-x$ is increasing in the interval:

• A. (0, 4)
• B. $(-\infty, 0)$
• C. (-1, 1)
• D. (-1, 0)
• E. $(0, \infty)$

Hint:

To find decreasing intervals, solve for $f'(x) < 0$.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
A function is increasing on an interval where its first derivative is positive. We need to find the derivative of the given function and then determine the interval for which this derivative is greater than zero.
Step 2: Key Formula or Approach:
1. Find the first derivative, \( f'(x) \).
2. Solve the inequality \( f'(x)>0 \) to find the interval(s) where the function is increasing.
Step 3: Detailed Explanation:
The function is \( f(x) = e^x - x \).
Step 3a: Find the first derivative
\[ f'(x) = \frac{d}{dx}(e^x - x) = \frac{d}{dx}(e^x) - \frac{d}{dx}(x) \] \[ f'(x) = e^x - 1 \] Step 3b: Solve the inequality \( f'(x)>0 \)
We need to find the values of x for which \( e^x - 1>0 \).
\[ e^x>1 \] To solve for x, we can think about the graph of \( y=e^x \). The value of \( e^x \) is 1 when \( x=0 \) (since \( e^0=1 \)). For any \( x>0 \), the value of \( e^x \) will be greater than 1.
Alternatively, we can take the natural logarithm of both sides. Since \( \ln(x) \) is an increasing function, the inequality direction is preserved.
\[ \ln(e^x)>\ln(1) \] \[ x>0 \] So, the function is increasing when \( x>0 \).
Step 4: Final Answer:
In interval notation, the function is increasing on \( (0, \infty) \).
We check the given options. Option (A) (0,4) is a sub-interval of \( (0, \infty) \). Option (E) \( (0, \infty) \) is the complete interval. In such cases, the most complete interval is the correct answer.