Question

The minimum of \(f(x)=\sqrt{(10-x^2)}\) in the interval \([-3,2]\) is

• A. \(\sqrt4\)
• B. \(\sqrt6\)
• C. \(1\)
• D. \(0\)
• E. \(\sqrt10\)

Answer 1

The Correct Option is C

Step 1: Identify the function and interval: \[ f(x) = \sqrt{10 - x^2} \] We need to find its minimum on \([-3, 2]\).

Step 2: Analyze the function behavior: The expression under the square root must be non-negative: \[ 10 - x^2 \geq 0 \implies x^2 \leq 10 \] This holds for all \(x \in [-3, 2]\).

Step 3: Find critical points by taking derivative: \[ f'(x) = \frac{d}{dx} \sqrt{10 - x^2} = \frac{-x}{\sqrt{10 - x^2}} \] Set \(f'(x) = 0\): \[ \frac{-x}{\sqrt{10 - x^2}} = 0 \implies x = 0 \]

Step 4: Evaluate function at critical point and endpoints: \[ f(-3) = \sqrt{10 - (-3)^2} = \sqrt{1} = 1 \] \[ f(0) = \sqrt{10 - 0^2} = \sqrt{10} \approx 3.162 \] \[ f(2) = \sqrt{10 - 2^2} = \sqrt{6} \approx 2.449 \]

Step 5: Determine the minimum: 

The smallest value among these is \(f(-3) = 1\).

Answer 2

Given
\(f(x) = √ (10 − x^2)\)  
So,
\(f(x) = \sqrt{(10 − x^2)}\)  is maximum when \(x^2\) is maximum.
Then, for [-3,2]
\(f(x) = √ (10 − x^2)\)  such that :
∴ minimum of \(f(x) = \sqrt{(10 − 9 )}\)
\(= \sqrt1 =1\)
So, the correct option is (C) : 1.