Check whether the following represent function or not
`x^(2)+y^(2)=36, xepsilon[0,1]`

Explanation

To determine whether the equation \( x^2 + y^2 = 36 \) represents a function with the given domain \( x \in [0, 1] \), we need to analyze the relationship between \( x \) and \( y \).

### Step 1: Understand the equation
The equation \( x^2 + y^2 = 36 \) represents a circle with a radius of 6 centered at the origin (0,0) in the Cartesian plane.

### Step 2: Solve for \( y \)
To check if it represents a function, we can solve for \( y \) in terms of \( x \):
\[ y^2 = 36 - x^2 \]
Taking the square root of both sides gives:
\[ y = \sqrt{36 - x^2} \quad \text{and} \quad y = -\sqrt{36 - x^2} \]

### Step 3: Analyze the outputs for a given input
Now, we need to check the outputs for specific values of \( x \) within the domain \( [0, 1] \).

- For \( x = 0 \):
\[ y = \sqrt{36 - 0^2} = \sqrt{36} = 6 \quad \text{and} \quad y = -\sqrt{36 - 0^2} = -6 \]
This gives two outputs: \( y = 6 \) and \( y = -6 \).

- For \( x = 1 \):
\[ y = \sqrt{36 - 1^2} = \sqrt{35} \quad \text{and} \quad y = -\sqrt{36 - 1^2} = -\sqrt{35} \]
Again, this gives two outputs: \( y = \sqrt{35} \) and \( y = -\sqrt{35} \).

### Step 4: Conclusion
Since for each input \( x \) in the domain \( [0, 1] \), there are two corresponding outputs for \( y \), this violates the definition of a function, which states that each input must have exactly one output.

Thus, we conclude that the equation \( x^2 + y^2 = 36 \) for \( x \in [0, 1] \) does **not** represent a function.

### Final Answer
The equation \( x^2 + y^2 = 36 \) for \( x \in [0, 1] \) is **not a function**.

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