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Problem Let f f f be a continuous functions satisfying
f ′ ( ln x ) = { 1 for 0 < x ≤ 1 e x − 1 for x > 1 f'(\ln x) = \begin{cases} 1 & \text{for } 0 < x \le 1 \\ e^x - 1 & \text{for } x > 1 \end{cases} f ′ ( ln x ) = { 1 e x − 1 for 0 < x ≤ 1 for x > 1
and f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 , then f ( x ) f(x) f ( x ) can be defined as
a. f ( x ) = { 1 , if x ≤ 0 1 − e x , if x > 0 f(x) = \begin{cases} 1, & \text{if } x \le 0 \\ 1 - e^x, & \text{if } x > 0 \end{cases} f ( x ) = { 1 , 1 − e x , if x ≤ 0 if x > 0
b. f ( x ) = { 1 , if x ≤ 0 e x − 1 , if x > 0 f(x) = \begin{cases} 1, & \text{if } x \le 0 \\ e^x - 1, & \text{if } x > 0 \end{cases} f ( x ) = { 1 , e x − 1 , if x ≤ 0 if x > 0
c. f ( x ) = { x , if x < 0 e x , if x > 0 f(x) = \begin{cases} x, & \text{if } x < 0 \\ e^x, & \text{if } x > 0 \end{cases} f ( x ) = { x , e x , if x < 0 if x > 0
d. f ( x ) = { x , if x ≤ 0 e x − 1 , if x > 0 f(x) = \begin{cases} x, & \text{if } x \le 0 \\ e^x - 1, & \text{if } x > 0 \end{cases} f ( x ) = { x , e x − 1 , if x ≤ 0 if x > 0
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