91Ó°ÊÓ

Solve the equation or inequality. $$\sqrt[3]{x} \leq x$$

Use \(f(x)=-2 x, g(x)=\sqrt{x}\) and \(h(x)=|x|\) to find and simplify expressions for the following functions and state the domain of each using interval notation. $$(g \circ h \circ f)(x)$$

Use \(f(x)=-2 x, g(x)=\sqrt{x}\) and \(h(x)=|x|\) to find and simplify expressions for the following functions and state the domain of each using interval notation. $$(f \circ h \circ g)(x)$$

With the help of your classmates, explain why a function which is either strictly increasing or strictly decreasing on its entire domain would have to be one-to-one, hence invertible.

Solve the equation or inequality. $$2(x-2)^{-\frac{1}{3}}-\frac{2}{3} x(x-2)^{-\frac{4}{3}} \leq 0$$

Use \(f(x)=-2 x, g(x)=\sqrt{x}\) and \(h(x)=|x|\) to find and simplify expressions for the following functions and state the domain of each using interval notation. $$(f \circ g \circ h)(x)$$

Solve the equation or inequality. $$-\frac{4}{3}(x-2)^{-\frac{4}{3}}+\frac{8}{9} x(x-2)^{-\frac{7}{3}} \geq 0$$

If \(f\) is odd and invertible, prove that \(f^{-1}\) is also odd.

Let \(f\) and \(g\) be invertible functions. With the help of your classmates show that \((f \circ g)\) is one-to-one, hence invertible, and that \((f \circ g)^{-1}(x)=\left(g^{-1} \circ f^{-1}\right)(x)\).

Solve the equation or inequality. $$2 x^{-\frac{1}{3}}(x-3)^{\frac{1}{3}}+x^{\frac{2}{3}}(x-3)^{-\frac{2}{3}} \geq 0$$