Week 6 Solutions

Answers for wk 6:

1) $x\in(-\infty,0)$
2) $\sqrt{x}$

Solution (1):

$x+\frac{1}{x}\leq -2$
$x+\frac{1}{x}+2\leq 0$
$\frac{x^2+2x+1}{x}\leq 0$
$\frac{(x+1)^2}{x}\leq 0$

Since the numerator of the fraction is always positive or 0, thus the expression on the LHS is only less than or equal to 0 if and only if the denominator, x, is negative. Hence, the range for x is actually:

$x\in(-\infty,0)$

Solution (2):

$\sqrt[3]{x\sqrt{x}}=\sqrt[3]{x^{\frac{3}{2}}}=x^{\frac{3}{6}}=x^{\frac{1}{2}}=\sqrt{x}$
$\Longrightarrow \sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}=\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}=\sqrt[3]{x\sqrt{x}}=\sqrt{x}$