4. Show that the equation $$ x^{3}+x^{2}-2 x-1=0 $$ has at least one solution in the interval $$ [1,3] $$.

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To show that the equation $$ x^{3}+x^{2}-2 x-1=0 $$ has at least one solution in the interval $$ [1,3] $$, we can use the Intermediate Value Theorem. This theorem states that if a continuous function changes sign over an interval, then there is at least one root in that interval.

1. **Define the function**:
   Let $$ f(x) = x^{3} + x^{2} - 2x - 1 $$.

2. **Evaluate the function at the endpoints of the interval**:
   - Calculate $$ f(1) $$:
     $$
     f(1) = 1^{3} + 1^{2} - 2 \cdot 1 - 1 = 1 + 1 - 2 - 1 = -1
     $$
   - Calculate $$ f(3) $$:
     $$
     f(3) = 3^{3} + 3^{2} - 2 \cdot 3 - 1 = 27 + 9 - 6 - 1 = 29
     $$

3. **Check the signs of $$ f(1) $$ and $$ f(3) $$**:
   - $$ f(1) = -1 $$ (negative)
   - $$ f(3) = 29 $$ (positive)

Since $$ f(1) < 0 $$ and $$ f(3) > 0 $$, the function $$ f(x) $$ changes sign over the interval $$ [1, 3] $$.

4. **Conclusion**:
By the Intermediate Value Theorem, since $$ f(x) $$ is continuous and changes sign between $$ x = 1 $$ and $$ x = 3 $$, there exists at least one solution $$ c $$ in the interval $$ [1, 3] $$ such that $$ f(c) = 0 $$.

Thus, we have shown that the equation $$ x^{3}+x^{2}-2 x-1=0 $$ has at least one solution in the interval $$ [1,3] $$.
The equation $$ x^{3}+x^{2}-2x-1=0 $$ has at least one solution between 1 and 3.

4. Show that the equation \( x^{3}+x^{2}-2 x-1=0 \) has at least one solution in the interval \( [1,3] \).

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