1. **Problem statement:** Given the function $f(x) = -x^5 + 2x^3 + x$ with $x \in \mathbb{R}$, prove the following statements: 2. **Symmetry of $G$ about the origin:** - A function is symmetric about the origin if it is an odd function, i.e., $f(-x) = -f(x)$ for all $x$. - Calculate $f(-x)$: $$f(-x) = -(-x)^5 + 2(-x)^3 + (-x) = -(-x^5) + 2(-x^3) - x = x^5 - 2x^3 - x$$ - Calculate $-f(x)$: $$-f(x) = -(-x^5 + 2x^3 + x) = x^5 - 2x^3 - x$$ - Since $f(-x) = -f(x)$, $f$ is an odd function. - Therefore, the graph $G$ is symmetric about the origin. 3. **Existence of at least one zero of $f$:** - Since $f$ is a polynomial function continuous on $\mathbb{R}$, and since $f(0) = 0$, the function has at least one zero at $x=0$. 4. **Existence of at least one local extremum:** - Local extrema occur where the first derivative $f'(x)$ is zero and the function changes direction. - Compute the derivative: $$f'(x) = \frac{d}{dx}(-x^5 + 2x^3 + x) = -5x^4 + 6x^2 + 1$$ - Since $f'(x)$ is continuous and a quartic polynomial, it can have up to 4 real roots. - Check values to confirm roots exist: - At $x=0$, $f'(0) = 1 > 0$ - At $x=1$, $f'(1) = -5 + 6 + 1 = 2 > 0$ - At $x=2$, $f'(2) = -5(16) + 6(4) + 1 = -80 + 24 + 1 = -55 < 0$ - Since $f'(1) > 0$ and $f'(2) < 0$, by the Intermediate Value Theorem, $f'(x)$ has at least one root between 1 and 2. - Similarly, checking other intervals shows at least one root exists. - Therefore, $f$ has at least one local extremum. **Final answers:** - 5.1: $G$ is symmetric about the origin because $f$ is an odd function. - 5.2: $f$ has at least one zero at $x=0$. - 5.3: $f$ has at least one local extremum because $f'(x)$ has at least one real root.