1. **Problem statement:** Determine if the function $g(f(x))$ where $f(x)=x^3+2$ and $g(x)=\cos x$ is even, odd, or neither. 2. **Recall definitions:** - A function $h(x)$ is **even** if $h(-x) = h(x)$ for all $x$. - A function $h(x)$ is **odd** if $h(-x) = -h(x)$ for all $x$. 3. **Evaluate $g(f(-x))$:** $$g(f(-x)) = g((-x)^3 + 2) = g(-x^3 + 2) = \cos(-x^3 + 2)$$ 4. **Evaluate $g(f(x))$:** $$g(f(x)) = g(x^3 + 2) = \cos(x^3 + 2)$$ 5. **Check if $g(f(x))$ is even:** Is $\cos(-x^3 + 2) = \cos(x^3 + 2)$? Since cosine is an even function, $\cos(-\theta) = \cos(\theta)$, but here the arguments differ by more than just a sign because of the $+2$ term. 6. **Check if $g(f(x))$ is odd:** Is $\cos(-x^3 + 2) = -\cos(x^3 + 2)$? No, cosine values are not negated by changing the sign inside the argument plus a constant. 7. **Conclusion:** Since neither condition holds, $g(f(x))$ is **neither even nor odd**. --- **Part b) Derivative of $\sqrt{2}$ from first principles:** 1. **Problem statement:** Find the derivative of the constant function $h(x) = \sqrt{2}$ using first principles. 2. **Recall definition of derivative from first principles:** $$h'(x) = \lim_{h \to 0} \frac{h(x+h) - h(x)}{h}$$ 3. **Apply to $h(x) = \sqrt{2}$:** $$h'(x) = \lim_{h \to 0} \frac{\sqrt{2} - \sqrt{2}}{h} = \lim_{h \to 0} \frac{0}{h} = 0$$ 4. **Explanation:** Since $\sqrt{2}$ is a constant, its derivative is zero everywhere. **Final answers:** - $g(f(x))$ is neither even nor odd. - The derivative of $\sqrt{2}$ is $0$.