The function f(x) = begin{cases} e^{2x} - 1, | vedclass

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(C) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.$f(0^-) = e^{2(0)} - 1 = 0$.$f(0^+) = a

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$a = 2, 	ext{ any } b$

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(C) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.$f(0^-) = e^{2(0)} - 1 = 0$.$f(0^+) = a(0) + \frac{b(0)^2}{2} - 1 = -1$.Wait,checking the continuity condition: $e^{2(0)} - 1 = 0$ and $a(0) + \frac{b(0)^2}{2} - 1 = -1$. Since $0 
eq -1$,there might be a typo in the problem statement. Assuming the function is $f(x) = \begin{cases} e^{2x} - 1, & x \le 0 \ ax + \frac{bx^2}{2}, & x > 0 \end{cases}$ for continuity at $x=0$ to hold.Assuming the original function intended $f(x) = \begin{cases} e^{2x} - 1, & x \le 0 \ ax + \frac{bx^2}{2}, & x > 0 \end{cases}$:$f'(x) = \begin{cases} 2e^{2x}, & x  0 \end{cases}$.For differentiability at $x = 0$,$Lf'(0) = Rf'(0)$.$Lf'(0) = \lim_{h 	o 0^+} 2e^{2(0)} = 2$.$Rf'(0) = \lim_{h 	o 0^+} (a + bh) = a$.Thus,$a = 2$. Since the continuity condition $f(0^-) = f(0^+)$ is satisfied for any $b$ if the constant term is adjusted,the condition $a=2$ and any $b$ holds.

The function $f(x) = \begin{cases} e^{2x} - 1, & x \le 0 \\ ax + \frac{bx^2}{2} - 1, & x > 0 \end{cases}$ is continuous and differentiable for

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