The function \(f(x) = x^2 - 2x\) is strictly decreasing in the interval.
- (-∞, -1)
- (-1, ∞)
- (-∞, 1)
- (1, ∞)
The Correct Option is C
Solution and Explanation
\(f(x) = x^2-2x\)
The function \(f(x) = x^2 - 2x\) is strictly decreasing if
\(f'(x) < 0\)
\(\frac {d}{dx}(x^2-2x) < 0\)
\(2x-2 < 0\)
\(2x < 2\)
\(x <1\)
\(⇒ x ∈ (-∞, 1)\)
So, the correct option is (C): \((-∞, 1)\)
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Concepts Used:
Increasing and Decreasing Functions
Increasing Function:
On an interval I, a function f(x) is said to be increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≤ f(y)
Decreasing Function:
On an interval I, a function f(x) is said to be decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≥ f(y)
Strictly Increasing Function:
On an interval I, a function f(x) is said to be strictly increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) < f(y)
Strictly Decreasing Function:
On an interval I, a function f(x) is said to be strictly decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) > f(y)
Graphical Representation of Increasing and Decreasing Functions
