Question:

Step 1: Determine the value of $f(0)$ Set $x=0$ and $y=0$ in the functional equation $f(x+y) = 2^x f(y) + 4^y f(x)$: $$f(0+0) = 2^0 f(0) + 4^0 f(0)$$ $$f(0) = 1 \cdot f(0) + 1 \cdot f(0)$$ $$f(0) = 2f(0)$$ This implies $f(0) = 0$. Step 2: Determine the general form of $f(x)$ Consider a solution of the form $f(x) = k(4^x - 2^x)$ for some constant $k$. Substitute this form into the functional equation: Left Hand Side (LHS): $$f(x+y) = k(4^{x+y} - 2^{x+y}) = k(4^x 4^y - 2^x 2^y)$$ Right Hand Side (RHS): $$2^x f(y) + 4^y f(x) = 2^x \cdot k(4^y - 2^y) + 4^y \cdot k(4^x - 2^x)$$ $$= k(2^x 4^y - 2^x 2^y + 4^y 4^x - 4^y 2^x)$$ $$= k(4^x 4^y - 2^x 2^y)$$ Since LHS = RHS, the form $f(x) = k(4^x - 2^x)$ is a valid solution to the functional equation. Step 3: Use the derivative condition $f'(0) = \ln 2$ to find the constant $k$ Differentiate $f(x) = k(4^x - 2^x)$ with respect to $x$: $$f'(x) = k(4^x \ln 4 - 2^x \ln 2)$$ Since $\ln 4 = \ln (2^2) = 2 \ln 2$, we have: $$f'(x) = k(4^x \cdot 2 \ln 2 - 2^x \ln 2)$$ Now, evaluate $f'(0)$: $$f'(0) = k(4^0 \cdot 2 \ln 2 - 2^0 \ln 2)$$ $$f'(0) = k(1 \cdot 2 \ln 2 - 1 \cdot \ln 2)$$ $$f'(0) = k(2 \ln 2 - \ln 2)$$ $$f'(0) = k \ln 2$$ Given $f'(0) = \ln 2$, we set the expressions equal: $$k \ln 2 = \ln 2$$ This implies $k=1$. Step 4: State the explicit form of $f(x)$ With $k=1$, the function is $f(x) = 1 \cdot (4^x - 2^x)$. $$f(x) = 4^x - 2^x$$ Step 5: Calculate $f(4)$ Substitute $x=4$ into the explicit form of $f(x)$: $$f(4) = 4^4 - 2^4$$ $$f(4) = 256 - 16$$ $$f(4) = 240$$ Step 6: Determine the minimum value of $f(x)$ Let $f(x) = 4^x - 2^x$. To find the minimum value, let $u = 2^x$. Since $x \in \mathbb{R}$, $u > 0$. The function can be rewritten as: $$f(x) = (2^x)^2 - 2^x = u^2 - u$$ To find the minimum value of this quadratic in $u$, we complete the square: $$u^2 - u = \left(u - \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2$$ $$u^2 - u = \left(u - \frac{1}{2}\right)^2 - \frac{1}{4}$$ The minimum value of this expression occurs when $u - \frac{1}{2} = 0$, i.e., $u = \frac{1}{2}$. Since $u = 2^x$, we have $2^x = \frac{1}{2} = 2^{-1}$, which means $x = -1$. The minimum value of $f(x)$ is: $$f_{min} = \left(\frac{1}{2} - \frac{1}{2}\right)^2 - \frac{1}{4} = 0 - \frac{1}{4} = -\frac{1}{4}$$ Step 7: Confirm the general form of $f(x)$ From Step 2 and Step 3, it was established that $f(x) = k(4^x - 2^x)$ for some constant $k$. With $k=1$, this form is consistent.