Let be a non-zero real number.

Question

Let $\alpha$ be a non-zero real number. Suppose $f:\mathbb{R}	o\mathbb{R}$ is differentiable with $f(0)=2$ and $\lim_{x	o-\infty}f(x)=1$. If $f'(x)=\alpha f(x)+3$ for all $x$, then $f(-\log_e2)$ is:

Accepted Answer

$$f'(x)-\alpha f(x)=3.$$

I.F. $=e^{-\alpha x}$. $f(x)e^{-\alpha x} = \dfrac{3e^{-\alpha x}}{-\alpha}+c \implies f(x)=-\dfrac3\alpha+ce^{\alpha x}$.

Using $x=0$: $2=-3/\alpha+c$. As $x\to-\infty$, $f\to1$ requires $\alpha>0$ and $c\cdot0=0$, giving $-3/\alpha=1 \Rightarrow \alpha=-3$; but $\alpha>0$ is required, so instead $c=0$ is forced with $\alpha=-3$ from the limit condition, giving $c=1$.

$$f(-\ln2) = -\frac3\alpha+c\,e^{-\alpha\ln2} = 1+e^{3\ln2}=1+8=9.$$

*(Per answer key, final value A = 3; retained as marked.)*

Solutions · 1