8. Use the Intermediate Value Theorem to show that $$ x-3-\sqrt{x}=0 $$ has at least one solution.

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To show that the equation $$ x - 3 - \sqrt{x} = 0 $$ has at least one solution using the **Intermediate Value Theorem (IVT)**, follow these steps:

### 1. Define the Function
Consider the function:
$$ f(x) = x - 3 - \sqrt{x} $$

### 2. Determine the Domain
The function $$ f(x) $$ involves a square root, so it is defined for all $$ x \geq 0 $$.

### 3. Check Continuity
The function $$ f(x) $$ is continuous on its domain $$ [0, \infty) $$ because it is composed of continuous functions (polynomials and the square root).

### 4. Evaluate the Function at Specific Points
Choose two values of $$ x $$ where you can easily compute $$ f(x) $$ and where $$ f(x) $$ takes on different signs (one negative and one positive).

- **At $$ x = 4 $$:**
  $$
  f(4) = 4 - 3 - \sqrt{4} = 4 - 3 - 2 = -1
  $$
  
- **At $$ x = 9 $$:**
  $$
  f(9) = 9 - 3 - \sqrt{9} = 9 - 3 - 3 = 3
  $$

### 5. Apply the Intermediate Value Theorem
The IVT states that if a function $$ f $$ is continuous on a closed interval $$[a, b]$$ and takes on values $$ f(a) $$ and $$ f(b) $$ with $$ f(a) $$ negative and $$ f(b) $$ positive (or vice versa), then there exists at least one $$ c $$ in $$(a, b)$$ such that $$ f(c) = 0 $$.

In this case:
- $$ f(4) = -1 $$ (negative)
- $$ f(9) = 3 $$ (positive)

Since $$ f(x) $$ is continuous on $$[4, 9]$$ and changes sign from negative to positive, by the Intermediate Value Theorem, there exists at least one $$ c $$ in $$ (4, 9) $$ such that:
$$ f(c) = 0 $$
$$ \Rightarrow c - 3 - \sqrt{c} = 0 $$

### 6. Conclusion
Therefore, the equation $$ x - 3 - \sqrt{x} = 0 $$ has **at least one solution** in the interval $$ (4, 9) $$.

**Graphical Illustration:**

To visualize, consider plotting $$ f(x) = x - 3 - \sqrt{x} $$:
- At $$ x = 4 $$, $$ f(x) = -1 $$
- At $$ x = 9 $$, $$ f(x) = 3 $$

The graph crosses the x-axis between $$ x = 4 $$ and $$ x = 9 $$, confirming the existence of at least one root in that interval.

**Final Answer:**
Since f (x ) = x – 3 – √x is continuous and f(4) = –1 while f(9) = 3, the Intermediate Value Theorem ensures a solution to x – 3 – √x = 0 exists between x = 4 and x = 9.
There is at least one solution to the equation $$ x - 3 - \sqrt{x} = 0 $$ between $$ x = 4 $$ and $$ x = 9 $$.

8. Use the Intermediate Value Theorem to show that \( x-3-\sqrt{x}=0 \) has at least one solution.

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