1. **State the problem:** We have two functions $f(x) = x^{3}$ and $g(x) = x^{2} + 1$. We want to determine which of the following functions are odd: (1) $f \times g$, meaning $(f \times g)(x) = f(x) \cdot g(x)$ (2) $f \circ g$, meaning $(f \circ g)(x) = f(g(x))$ (3) $g \circ f$, meaning $(g \circ f)(x) = g(f(x))$ 2. **Recall definitions:** - A function $h$ is odd if $h(-x) = -h(x)$ for all $x$. - $f(x) = x^{3}$ is an odd function because $f(-x) = (-x)^{3} = -x^{3} = -f(x)$. - $g(x) = x^{2} + 1$ is not odd because $g(-x) = (-x)^{2} + 1 = x^{2} + 1 = g(x)$, so it is even. 3. **Check (1) $f \times g$:** $$(f \times g)(x) = f(x) \cdot g(x) = x^{3} (x^{2} + 1) = x^{5} + x^{3}$$ Check if odd: $$(f \times g)(-x) = (-x)^{5} + (-x)^{3} = -x^{5} - x^{3} = -(x^{5} + x^{3}) = -(f \times g)(x)$$ So, $(f \times g)$ is odd. 4. **Check (2) $f \circ g$:** $$(f \circ g)(x) = f(g(x)) = (x^{2} + 1)^{3}$$ Check if odd: $$(f \circ g)(-x) = ( (-x)^{2} + 1 )^{3} = (x^{2} + 1)^{3} = (f \circ g)(x)$$ Since $(f \circ g)(-x) = (f \circ g)(x)$, it is even, not odd. 5. **Check (3) $g \circ f$:** $$(g \circ f)(x) = g(f(x)) = (x^{3})^{2} + 1 = x^{6} + 1$$ Check if odd: $$(g \circ f)(-x) = ((-x)^{3})^{2} + 1 = (-x^{3})^{2} + 1 = x^{6} + 1 = (g \circ f)(x)$$ So, $(g \circ f)$ is even, not odd. 6. **Conclusion:** Only $(f \times g)$ is odd. **Final answer:** (a) (1) only