Prime equals triple prime?

Integrate $f'(x) = f'''(x)$ to see $f(x) = f''(x) + C$ The solution to the differential equation $f(x) = f''(x)$ is well known to be $f(x) = A\times e^x + B\times e^{-x},$ but if you want a proof, I recommend studying second order homogeneous differential equations with constant coefficients. Now the solution to our original equation is the same as this one, but with an additional constant of integration. Therefore the solution is of the form $f(x) = Ae^x + Be^{-x} + C$ You can find the first and second derivatives of this function, then set up a system of equations using the given values in the problem to solve for the constants. You should find $f(x) = \frac{1}{e}\times e^x - 4e\times e^{-x} + 9$ Substitute $x = \ln(42).$ $f(\ln(42)) \approx \boxed{24.192}$

Assume that $f(x)$ takes the form of $f(x) = Ae^{kx}+Be^{-kx}+C$ , then we have:

$\begin{cases} f(x) & = Ae^{kx} & +Be^{-kx} & +C & ...(1)\\ f'(x) & = Ake^{kx} & -Bke^{-kx} & & ...(2)\\ f''(x) & = Ak^2e^{kx} & +Bk^2e^{-kx} & & ...(3)\\ f'''(x) & = Ak^3e^{kx} &- Bk^3e^{-kx} & & ...(4) \end{cases}$

For $f'(x) = f'''(x) \quad \Rightarrow k = 1$ .

Therefore:

$\begin{cases} f(1) & = Ae +Be^{-1} +C & = 6 & ...(1a)\\ f'(1) & = Ae -Be^{-1} &= 5 & ...(2a)\\ f''(1) & = Ae +Be^{-1} & = -3 & ...(3a) \end{cases}$

$\begin{cases} Eq.2a+Eq.3: & 2Ae = 2 & \Rightarrow A = e^{-1} \\ Eq.2a: & 1 - Be^{-1} = 5 & \Rightarrow B = -4e \\ Eq.1a+Eq.3a: & C = 6-(-3) & \Rightarrow C = 9 \end{cases}$

$\Rightarrow f(x) = e^{x-1} - 4e^{-(x-1)} + 9$

$\Rightarrow f \left( \ln{(42)} \right) =\dfrac {42}{e}- \dfrac{4e}{42}+ 9 = \boxed{24.192}$