1. **Problem statement:** Find possible function equations for the given graphs based on their properties. 2. **Part a) cubic function with local max and min:** - Given points: local max at $H(-1,1)$, local min at $T(1,-1)$, and $f(0)=0$. - Since it has one local max and one local min, the function is cubic: $f(x)=ax^3+bx^2+cx+d$. - The function is odd symmetric because $f(0)=0$ and the max and min are symmetric about the origin, so $f(-x)=-f(x)$. - This implies $b=0$ and $d=0$, so $f(x)=ax^3+cx$. 3. **Use conditions to find $a$ and $c$:** - Derivative: $f'(x)=3ax^2+c$. - Local extrema at $x=-1$ and $x=1$ means $f'(-1)=0$ and $f'(1)=0$. - $f'(1)=3a(1)^2+c=3a+c=0$ and $f'(-1)=3a(1)^2+c=3a+c=0$ (same equation). - So $c=-3a$. 4. **Use function values at extrema:** - $f(-1)=a(-1)^3+c(-1)=-a - c=1$. - Substitute $c=-3a$: $-a - (-3a)= -a + 3a=2a=1$ so $a=\frac{1}{2}$. - Then $c=-3 \times \frac{1}{2}=-\frac{3}{2}$. 5. **Final function for a):** $$f(x)=\frac{1}{2}x^3 - \frac{3}{2}x$$ --- 6. **Part b) quartic even function:** - Symmetric about y-axis, so $f(x)=ax^4+bx^2+c$. - Local minima at $T_1(-2,-4)$ and $T_2(2,-4)$, local max at origin $(0,0)$. 7. **Use $f(0)=c=0$** - So $f(x)=ax^4+bx^2$. 8. **Find derivative:** $$f'(x)=4ax^3 + 2bx$$ - Set $f'(0)=0$ (max at 0), and $f'(\pm 2)=0$ (minima). 9. **Solve $f'(2)=0$: $$4a(2)^3 + 2b(2) = 32a + 4b = 0 \Rightarrow 8a + b = 0 \Rightarrow b = -8a$$** 10. **Use $f(2) = a(2)^4 + b(2)^2 = 16a + 4b = -4$** - Substitute $b=-8a$: $16a + 4(-8a) = 16a - 32a = -16a = -4$ so $a=\frac{1}{4}$. - Then $b = -8 \times \frac{1}{4} = -2$. 11. **Final function for b):** $$f(x) = \frac{1}{4}x^4 - 2x^2$$ --- 12. **Part c) higher-degree even function:** - Even function with global min at $T(0,0)$ and local maxima at $W_1(-2,20)$ and $W_2(2,20)$. - Assume degree 6 polynomial: $f(x) = ax^6 + bx^4 + cx^2 + d$. - Even symmetry implies only even powers. 13. **Use $f(0) = d = 0$** - So $f(x) = ax^6 + bx^4 + cx^2$. 14. **Use $f(2) = 64a + 16b + 4c = 20$** 15. **Derivative:** $$f'(x) = 6ax^5 + 4bx^3 + 2cx$$ - At $x=0$, $f'(0)=0$ (min). - At $x=2$, $f'(2) = 192a + 32b + 4c = 0$ (max). 16. **System of equations:** $$\begin{cases} 64a + 16b + 4c = 20 \\ 192a + 32b + 4c = 0 \end{cases}$$ 17. **Subtract first from second:** $$(192a - 64a) + (32b - 16b) + (4c - 4c) = 0 - 20$$ $$128a + 16b = -20$$ 18. **Divide by 4:** $$32a + 4b = -5$$ 19. **Express $b$:** $$4b = -5 - 32a \Rightarrow b = \frac{-5 - 32a}{4}$$ 20. **Substitute $b$ into first equation:** $$64a + 16 \times \frac{-5 - 32a}{4} + 4c = 20$$ $$64a + 4(-5 - 32a) + 4c = 20$$ $$64a - 20 - 128a + 4c = 20$$ $$-64a - 20 + 4c = 20$$ 21. **Solve for $c$:** $$4c = 20 + 20 + 64a = 40 + 64a \Rightarrow c = 10 + 16a$$ 22. **Choose $a$ to simplify, e.g., $a=0$:** - Then $b = \frac{-5 - 0}{4} = -\frac{5}{4}$ - $c = 10 + 0 = 10$ 23. **Final function for c):** $$f(x) = 0 \cdot x^6 - \frac{5}{4}x^4 + 10x^2 = -\frac{5}{4}x^4 + 10x^2$$ This matches the shape with maxima at $x=\pm 2$ and minimum at 0. --- **Summary:** - a) $f(x) = \frac{1}{2}x^3 - \frac{3}{2}x$ - b) $f(x) = \frac{1}{4}x^4 - 2x^2$ - c) $f(x) = -\frac{5}{4}x^4 + 10x^2$