Find the values of x for which \(y=[x(x-2)]^2\) is an increasing function.
Find the values of x for which \(y=[x(x-2)]^2\) is an increasing function.
Solution and Explanation
We have,
y = [x(x-2)]2 = [x2-2x]2
\(\frac {dy}{dx}\) = y'(x2-2x) = 4x(x-2)(x-1)
\(\frac {dy}{dx}\) = 0\(\implies\)x=0, x=2, x=1.
The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals i.e.,
(-∞,0),(0,1)(1,2), and (2,∞).
In intervals (-∞,0) and (1,2), \(\frac {dy}{dx}\)<0
∴ y is strictly decreasing in intervals (-∞,0) and (1,2).
However, in intervals (0, 1) and (2, ∞), \(\frac {dy}{dx}\)>0.
∴ y is strictly increasing in intervals (0,1) and (2,∞).
y is strictly increasing for 0<x<1 and x>2.
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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
