Prove that the function f given by \(f(x) = x^2 − x + 1\) is neither strictly increasing nor strictly decreasing,on \((−1, 1)\).
Prove that the function f given by \(f(x) = x^2 − x + 1\) is neither strictly increasing nor strictly decreasing,on \((−1, 1)\).
Solution and Explanation
The given function is f(x) = x2 − x + 1.
f'(x) = 2x-1
Now, f'(x) = 0\(\implies\) x = \(\frac 12\).
The point \(\frac 12\)divides the interval (−1, 1) into two disjoint intervals
i.e., (-1,\(\frac 12\)) and (\(\frac 12\),1).
Now, in interval (-1,\(\frac 12\)), f'(x) = 2x-1<0.
Therefore, f is strictly decreasing in interval (-1,\(\frac 12\)).
However, in interval (\(\frac 12\),1), f'(x) = 2x-1>0.
Therefore, f is strictly increasing in interval (\(\frac 12\),1).
Hence, f is neither strictly increasing nor decreasing in interval (−1, 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
