1. **Problem statement:** We have a cubic polynomial $p(x)$ such that for $j=1,2,3,4$, $p(\frac{1}{j})=2j-1$. We need to find $p(-1)$. 2. **Understanding the problem:** Since $p(x)$ is cubic, it can be written as $p(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. 3. **Given conditions:** For $j=1,2,3,4$, $$p\left(\frac{1}{j}\right)=2j-1.$$ This gives four equations: $$p(1)=2(1)-1=1,$$ $$p\left(\frac{1}{2}\right)=2(2)-1=3,$$ $$p\left(\frac{1}{3}\right)=2(3)-1=5,$$ $$p\left(\frac{1}{4}\right)=2(4)-1=7.$$ 4. **Define a new polynomial:** Let $$q(x)=p(x)-\frac{2}{x}+1.$$ Note that for $x=\frac{1}{j}$, $$q\left(\frac{1}{j}\right)=p\left(\frac{1}{j}\right)-2j+1=0.$$ So $q(x)$ has roots at $x=1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$. 5. **Degree considerations:** Since $p(x)$ is cubic, $q(x)$ is $p(x)-\frac{2}{x}+1$. The term $\frac{2}{x}$ is not polynomial, but multiplying by $x$ gives a polynomial: $$xq(x)=xp(x)-2+ x.$$ Since $p(x)$ is degree 3, $xp(x)$ is degree 4, so $xq(x)$ is degree 4 polynomial with roots at $x=1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$. 6. **Form of $xq(x)$:** $$xq(x)=k(x-1)(x-\frac{1}{2})(x-\frac{1}{3})(x-\frac{1}{4})$$ for some constant $k$. 7. **Rewrite $q(x)$:** $$q(x)=\frac{k(x-1)(x-\frac{1}{2})(x-\frac{1}{3})(x-\frac{1}{4})}{x}.$$ 8. **Recall $q(x)=p(x)-\frac{2}{x}+1$, so:** $$p(x)=q(x)+\frac{2}{x}-1=\frac{k(x-1)(x-\frac{1}{2})(x-\frac{1}{3})(x-\frac{1}{4})}{x}+\frac{2}{x}-1.$$ 9. **Since $p(x)$ is cubic, the numerator of $p(x)$ must be degree 3:** Multiply both sides by $x$: $$xp(x)=k(x-1)(x-\frac{1}{2})(x-\frac{1}{3})(x-\frac{1}{4}) + 2 - x.$$ The right side is degree 4 polynomial plus constants, so for $xp(x)$ to be degree 4 polynomial, the degree 4 term must vanish. 10. **Find $k$ by eliminating the $x^4$ term:** Expand leading term of $(x-1)(x-\frac{1}{2})(x-\frac{1}{3})(x-\frac{1}{4})$ is $x^4$. So leading term of right side is $k x^4$. Since $xp(x)$ is degree 4 polynomial, and $p(x)$ is cubic, $xp(x)$ has degree 4. But $xp(x)$ must be polynomial, so the $x^4$ term must be zero to keep $p(x)$ cubic. Therefore, $k=0$. 11. **If $k=0$, then:** $$xp(x)=2 - x,$$ so $$p(x)=\frac{2 - x}{x} = \frac{2}{x} - 1,$$ which is not a polynomial. This contradicts the assumption that $p(x)$ is a cubic polynomial. 12. **Reconsider approach:** Instead, define $$r(j) = p\left(\frac{1}{j}\right) = 2j - 1,$$ which is a linear function in $j$. Since $p(x)$ is cubic, and $p(1/j)$ is linear in $j$, try expressing $p(x)$ in terms of $x$: Let $p(x) = A + Bx + Cx^2 + Dx^3$. Given $p(1/j) = 2j - 1$, substitute $x=1/j$: $$p\left(\frac{1}{j}\right) = A + B\frac{1}{j} + C\frac{1}{j^2} + D\frac{1}{j^3} = 2j - 1.$$ Multiply both sides by $j^3$: $$A j^3 + B j^2 + C j + D = (2j - 1) j^3 = 2 j^4 - j^3.$$ 13. **Rewrite:** $$A j^3 + B j^2 + C j + D = 2 j^4 - j^3.$$ 14. **Equate coefficients:** The left side is degree 3 polynomial in $j$, right side degree 4. For equality for all $j$, coefficients of $j^4$ must be zero on left side, so no $j^4$ term on left side. Contradiction unless $p(x)$ is not cubic. 15. **Conclusion:** The problem as stated is inconsistent if $p(x)$ is cubic polynomial. 16. **Alternative approach:** Since $p(1/j) = 2j - 1$, define a polynomial $Q(j) = p(1/j) - (2j - 1)$ which has roots at $j=1,2,3,4$. Since $Q(j)$ has 4 roots, $Q(j)$ is at least degree 4 polynomial in $j$. But $p(1/j)$ is a rational function in $j$, so $p(x)$ must be of form: $$p(x) = \frac{a x^3 + b x^2 + c x + d}{x^3}$$ which is not polynomial. 17. **Given the problem, the only consistent polynomial $p(x)$ satisfying $p(1/j) = 2j - 1$ for $j=1,2,3,4$ is:** $$p(x) = -2 + \frac{3}{x}$$ which is not cubic polynomial. 18. **Therefore, the problem likely means $p(x)$ is cubic polynomial in $x$ satisfying $p(1/j) = 2j - 1$ for $j=1,2,3,4$. We can find $p(-1)$ by interpolation:** 19. **Use Lagrange interpolation for points:** $$(x_i, y_i) = \left(1,1\right), \left(\frac{1}{2},3\right), \left(\frac{1}{3},5\right), \left(\frac{1}{4},7\right).$$ 20. **Lagrange basis polynomials:** $$L_i(x) = \prod_{j \neq i} \frac{x - x_j}{x_i - x_j}.$$ 21. **Calculate $p(-1)$:** $$p(-1) = \sum_{i=1}^4 y_i L_i(-1).$$ 22. **Calculate each $L_i(-1)$:** - $L_1(-1) = \frac{(-1 - \frac{1}{2})(-1 - \frac{1}{3})(-1 - \frac{1}{4})}{(1 - \frac{1}{2})(1 - \frac{1}{3})(1 - \frac{1}{4})} = \frac{(-\frac{3}{2})(-\frac{4}{3})(-\frac{5}{4})}{(\frac{1}{2})(\frac{2}{3})(\frac{3}{4})} = \frac{-\frac{3}{2} \times -\frac{4}{3} \times -\frac{5}{4}}{\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}} = \frac{-\frac{5}{2}}{\frac{1}{4}} = -10.$$ - $L_2(-1) = \frac{(-1 - 1)(-1 - \frac{1}{3})(-1 - \frac{1}{4})}{(\frac{1}{2} - 1)(\frac{1}{2} - \frac{1}{3})(\frac{1}{2} - \frac{1}{4})} = \frac{(-2)(-\frac{4}{3})(-\frac{5}{4})}{(-\frac{1}{2})(\frac{1}{6})(\frac{1}{4})} = \frac{-\frac{10}{3}}{-\frac{1}{48}} = 160.$$ - $L_3(-1) = \frac{(-1 - 1)(-1 - \frac{1}{2})(-1 - \frac{1}{4})}{(\frac{1}{3} - 1)(\frac{1}{3} - \frac{1}{2})(\frac{1}{3} - \frac{1}{4})} = \frac{(-2)(-\frac{3}{2})(-\frac{5}{4})}{(-\frac{2}{3})(-\frac{1}{6})(\frac{1}{12})} = \frac{-\frac{15}{4}}{\frac{1}{36}} = -135.$$ - $L_4(-1) = \frac{(-1 - 1)(-1 - \frac{1}{2})(-1 - \frac{1}{3})}{(\frac{1}{4} - 1)(\frac{1}{4} - \frac{1}{2})(\frac{1}{4} - \frac{1}{3})} = \frac{(-2)(-\frac{3}{2})(-\frac{4}{3})}{(-\frac{3}{4})(-\frac{1}{4})(-\frac{1}{12})} = \frac{-4}{-\frac{1}{48}} = -192.$$ 23. **Calculate $p(-1)$:** $$p(-1) = 1 \times (-10) + 3 \times 160 + 5 \times (-135) + 7 \times (-192) = -10 + 480 - 675 - 1344 = -1549.$$