Question

Let $f(x) = \begin{cases} e^{x-1}, & x<0 \\ x^2 - 5x + 6, & x \ge 0 \end{cases}$ and $g(x) = f(|x|) + |f(x)|$. If the number of points where $g$ is not continuous and is not differentiable are $\alpha$ and $\beta$ respectively, then $\alpha + \beta$ is equal to _______.

Answer 1

Correct Answer: 4

To find where the function \( g(x) = f(|x|) + |f(x)| \) is not continuous or not differentiable, we first examine the piecewise function \( f(x) \).
Step 1: Analyze \( f(x) \)
▪ For \( x < 0 \), \( f(x) = e^{x-1} \). 
▪ For \( x \ge 0 \), \( f(x) = x^2 - 5x + 6 \).
Step 2: Examine \( g(x) = f(|x|) + |f(x)| \)
\( g(x) \) depends on both \( f(|x|) \) and \( |f(x)| \), adjusting \( f \) based on absolute values:
▪ For \( x \ge 0 \), \( f(|x|) = x^2 - 5x + 6 \) and \(|f(x)| = |x^2 - 5x + 6|\).
▪ For \( x < 0 \), \( f(|x|) = e^{-x-1} \) and \( |f(x)| = |e^{x-1}|\).
Step 3: Check continuity at \( x = 0 \)
\( f \) changes its form here, hence 
\(\lim_{x \to 0^-} f(x) = e^{-1}\), \(\lim_{x \to 0^+} f(x) = 6\).
Since the values do not equal, \( f(x) \) is not continuous at \( x = 0 \).
Thus, \( g(x) \) is not continuous at \( x = 0 \). \((\alpha = 1)\).
Step 4: Check differentiability at \( x = 0 \)
For differentiability, both sides must align in the first derivative.
▪ Derivative for \( x < 0 \): \( f'(x) = e^{x-1} \).
▪ Derivative for \( x \ge 0 \): \( f'(x) = 2x - 5 \).
\(\lim_{x \to 0^-} f'(x) = e^{-1}, \lim_{x \to 0^+} f'(x) = -5\).
These are not equal, hence not differentiable at \( x = 0 \).
\( g(x) \) is not differentiable at \( x = 0 \). \((\beta = 1)\).
Conclusion: \(\alpha + \beta = 2\).
\( 2 \) falls within the prescribed range \( [4,4] \).