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Question

Let $g(x)$ be a linear function and $f(x)=\left\{\begin{array}{cl}g(x) & , x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & , x>0\end{array}\right.$, is continuous at $x=0$. If $f^{\prime}(1)=f(-1)$, then the value $g(3)$ is

$\log _e\left(\frac{4}{9}\right)-1$
$\frac{1}{3} \log _e\left(\frac{4}{9 e^{1 / 3}}\right)$
$\log _e\left(\frac{4}{9 e^{1 / 3}}\right)$
$\frac{1}{3} \log _e\left(\frac{4}{9}\right)+1$

Solution

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