Question

The function f(x) = x² - 2x is strictly decreasing in the interval

• A. (-∞, 1)
• B. (1, ∞)
• C. R
• D. (-∞, ∞)

Answer 1

The Correct Option is A

We are given the function \( f(x) = x^2 - 2x \).
Step 1: Find the derivative of \( f(x) \)}
The derivative of the function is: \[ f'(x) = \frac{d}{dx} (x^2 - 2x) = 2x - 2 \]
Step 2: Find where \( f'(x) < 0 \)}
For the function to be strictly decreasing, we need: \[ f'(x) < 0 \] This gives: \[ 2x - 2 < 0 \] Solving for \( x \): \[ 2x < 2 \] \[ x < 1 \] Thus, the function is strictly decreasing in the interval \( (-\infty, 1) \).

Therefore, the correct answer is (A) \( (-\infty, 1) \).

Answer 2

Given:
Function: $f(x) = x^2 - 2x$ 

Step 1: Find the derivative
To determine where the function is increasing or decreasing, find the first derivative:
$$ f'(x) = \frac{d}{dx}(x^2 - 2x) = 2x - 2 $$ 
Step 2: Find critical point
Set $f'(x) = 0$ to find where the slope is zero:
$$ 2x - 2 = 0 \Rightarrow x = 1 $$ 
Step 3: Test intervals around the critical point
- For $x < 1$, choose $x = 0$: $f'(0) = 2(0) - 2 = -2 \Rightarrow$ Negative slope (decreasing) - For $x > 1$, choose $x = 2$: $f'(2) = 2(2) - 2 = 2 \Rightarrow$ Positive slope (increasing) 
Conclusion:
The function is strictly decreasing in the interval $(-\infty, 1)$