Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 223: 41

$f(x)=\dfrac{1}{2}e^{2x}+1$

Here, the derivative is given as: $f'(x)=e^{2x}$, thus, the general form of the function is: $f(x)=\dfrac{1}{2}e^{2x}+C$ The graph of the given function passes through $(0,\dfrac{3}{2})$ Therefore, $f(0)=\dfrac{3}{2} \implies \dfrac{3}{2}=\dfrac{1}{2}e^{0}+C$ Thus, $C=1$ Hence, $f(x)=\dfrac{1}{2}e^{2x}+1$