Answer: f(x) = cosh(x)
Step-by-step explanation:
[tex]f'(x)=\int\ f''{(x)} \, dx \\\\.\qquad =\int\ \dfrac{1}{2}(e^x+e^{-x})\ dx\\\\.\qquad =\dfrac{1}{2}\int (e^x+e^{-x})\ dx\\\\.\qquad =\dfrac{1}{2}\int 2\ cosh(x)\ dx\\\\.\qquad =\dfrac{1}{2}\cdot 2 \int cosh(x)\ dx\\\\.\qquad =\int cosh(x)\ dx\\\\.\qquad =sinh(x)+ C\\\\\text{Find the constant C by using the given condition that f'(0) = 0}\\0=sinh(0)+C\\0 = 0 + C\\0=C\\\\\text{So, f'(x) = sinh(x)}[/tex]
[tex]f(x)=\int f'(x)\ dx\\\\.\qquad =\int sinh(x)\ dx\\\\.\qquad =cosh(x)+C\\\\\text{Find the constant C by using the given condition that f(0) = 1}\\1=cosh(0)+C\\1 = 1 + C\\0=C\\\\\text{So, f(x) = cosh(x)}[/tex]
