If f(x) is differentiable on R,f(x) f^{prime} | vedclass

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(B) Given the differential equation $f(x) f^{\prime}(-x) - f(-x) f^{\prime}(x) = 0$. This can be rewritten as $\frac{d}{dx} [f(x) f(-x)] = f(x) f^{\pr

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Given the differential equation $f(x) f^{\prime}(-x) - f(-x) f^{\prime}(x) = 0$. This can be rewritten as $\frac{d}{dx} [f(x) f(-x)] = f(x) f^{\prime}(-x) (-1) + f^{\prime}(x) f(-x) = 0$. Wait,let us re-evaluate: $\frac{d}{dx} [f(x) f(-x)] = f^{\prime}(x) f(-x) + f(x) f^{\prime}(-x) (-1) = f^{\prime}(x) f(-x) - f(x) f^{\prime}(-x) = 0$. Since $f(x) f^{\prime}(-x) - f(-x) f^{\prime}(x) = 0$,we have $f^{\prime}(x) f(-x) - f(x) f^{\prime}(-x) = 0$. Thus,$\frac{d}{dx} [f(x) f(-x)] = 0$. This implies $f(x) f(-x) = c$,where $c$ is a constant. At $x = 0$,$f(0) f(0) = c \Rightarrow 3 	imes 3 = 9$,so $c = 9$. Thus,$f(x) f(-x) = 9$ for all $x \in R$. For $x = 3$,$f(3) f(-3) = 9 \Rightarrow 9 	imes f(-3) = 9 \Rightarrow f(-3) = 1$. Finally,$(1 + f(-3))^3 + 1 = (1 + 1)^3 + 1 = 2^3 + 1 = 8 + 1 = 9$.

If $f(x)$ is differentiable on $R$,$f(x) f^{\prime}(-x) - f(-x) f^{\prime}(x) = 0$,$f(0) = 3$,and $f(3) = 9$,then $(1 + f(-3))^3 + 1 = $

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