Let's solve the problem step by step using the **Intermediate Value Theorem (IVT)**. --- ### **Step 1: Define the Function** Let \( f(x) = x^3 + 2x - 5 \). --- ### **Step 2: Check Continuity** \( f(x) \) is a polynomial, so it is continuous everywhere, including on the interval \([0, 2]\). --- ### **Step 3: Evaluate \( f(x) \) at the Endpoints** Calculate \( f(0) \) and \( f(2) \): - \( f(0) = (0)^3 + 2 \cdot 0 - 5 = -5 \) - \( f(2) = (2)^3 + 2 \cdot 2 - 5 = 8 + 4 - 5 = 7 \) --- ### **Step 4: Check for a Sign Change** - \( f(0) = -5 \) (negative) - \( f(2) = 7 \) (positive) There is a sign change from negative to positive as \( x \) goes from 0 to 2. --- ### **Step 5: Apply the Intermediate Value Theorem** The IVT states: >If \( f \) is continuous on \([a,b]\) and \( N \) is any number between \( f(a) \) and \( f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f(c) = N \). Here: - \( f(0) = -5 < 0 \) - \( f(2) = 7 > 0 \) So, 0 is between \( f(0) \) and \( f(2) \). --- ### **Step 6: Conclusion** By the Intermediate Value Theorem, **there exists at least one \( c \) in the interval \((0, 2)\) such that \( f(c) = 0 \)**. This proves the equation \( x^3 + 2x - 5 = 0 \) has at least one root in \([0, 2]\). --- ## **Final Answer** > By the Intermediate Value Theorem, the equation \( x^3 + 2x - 5 = 0 \) has at least one root in the interval \([0, 2]\) because the function changes sign between \( x = 0 \) and \( x = 2 \), and is continuous on that interval.

Let's provide a detailed step-by-step solution with clear calculations and explanations for using the **Intermediate Value Theorem (IVT)** to show that the equation \[ x^3 + 2x - 5 = 0 \] has at least one root in the interval \([0, 2]\). --- ### **Step 1: Define the function** Let: \[ f(x) = x^3 + 2x - 5 \] Our goal is to show that there exists some \( c \in [0, 2] \) such that \( f(c) = 0 \). --- ### **Step 2: Verify the continuity of \(f(x)\)** Since \(f(x)\) is a polynomial function, it is **continuous everywhere** (including on \([0, 2]\)). This is essential because the IVT applies only to continuous functions. --- ### **Step 3: Evaluate \(f(x)\) at the endpoints of the interval** Calculate \(f(0)\): \[ f(0) = (0)^3 + 2 \times 0 - 5 = 0 + 0 - 5 = -5 \] Calculate \(f(2)\): \[ f(2) = (2)^3 + 2 \times 2 - 5 = 8 + 4 - 5 = 7 \] **Summary:** | \( x \) | \( f(x) \) | |---------|------------| | 0 | -5 | | 2 | 7 | --- ### **Step 4: Check for a sign change** Observe that: - \(f(0) = -5\) (negative) - \(f(2) = 7\) (positive) Since \(f(0)\) is negative and \(f(2)\) is positive, the function changes sign over \([0, 2]\). --- ### **Step 5: Apply the Intermediate Value Theorem** The **IVT states**: > If \(f\) is continuous on \([a, b]\), then for **any** value \(N\) between \(f(a)\) and \(f(b)\), there exists some \( c \in (a, b) \) such that \(f(c) = N\). Here: - \(f(0) = -5\) - \(f(2) = 7\) Since 0 (the target value for the root) lies **between** \(-5\) and 7, the IVT guarantees **at least one \(c \in (0, 2)\)** such that: \[ f(c) = 0 \] --- ### **Step 6: Final conclusion** Because \(f(x)\) is continuous on \([0, 2]\), and \(f(0) < 0\), \(f(2) > 0\), the IVT ensures that **there exists at least one \( c \in (0, 2) \)** such that: \[ x^3 + 2x - 5 = 0 \] --- ## **Final Answer:** **Using the Intermediate Value Theorem, we have shown that the polynomial \(f(x) = x^3 + 2x - 5\) has at least one root in the interval \([0, 2]\). This is because \(f(0) = -5 < 0\) and \(f(2) = 7 > 0\), and the function is continuous on \([0, 2]\). Therefore, there exists some \( c \in (0, 2) \) such that \( f(c) = 0 \).**

The Intermediate Value Theorem states that if a function \(f\) is continuous on a closed interval \([a, b]\), then it takes on every value between \(f(a)\) and \(f(b)\) at some point within \([a, b]\). Since \(f(x) = x^3 + 2x - 5\) is continuous everywhere, it must have at least one root in \([0, 2]\) because \(f(0) < 0\) and \(f(2) > 0\).