Mathematics - Continuity - Differentiability Question with Solution | TestHub

The functional equation $f(x+y)=f(x)f(y)$ for a differentiable function implies $f(x)=e^{kx}$ for some constant $k$. Differentiating $f(x)=e^{kx}$ gives $f^{\prime }(x)=k{e}^{kx}$. Given $f^{\prime }(0)=3$, we have $k{e}^0=k=3$. So $f(x)=e^{3x}$. For $g(x)$ to be continuous at $x=0$, $M={\lim }_{x\to 0}g(x)={\lim }_{x\to 0}\frac{e^{3x}-1-3x}{x^2}$. This is an indeterminate form $\frac{0}{0}$. Applying L'Hopital's Rule twice: $M={\lim }_{x\to 0}\frac{3e^{3x}-3}{2x}={\lim }_{x\to 0}\frac{9e^{3x}}{2}=\frac{9e^0}{2}=\frac{9}{2}$. Thus, $M=\frac{9}{2}$.