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Question
Show that the function x2 − x + 1 is neither increasing nor decreasing on (0, 1) ?
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Solution
\[f\left( x \right) = x^2 - x + 1\]
\[f'\left( x \right) = 2x - 1\]
\[\text { For f(x) to be increasing, we must have }\]
\[f'\left( x \right) > 0\]
\[ \Rightarrow 2x - 1 > 0\]
\[ \Rightarrow 2x > 1\]
\[ \Rightarrow x > \frac{1}{2}\]
\[ \Rightarrow x \in \left( \frac{1}{2}, 1 \right)\]
\[\text { So,f(x)is increasing on }\left( \frac{1}{2}, 1 \right) . \]
\[\text{ For f(x) to be decreasing, we must have }\]
\[f'\left( x \right) < 0\]
\[ \Rightarrow 2x - 1 < 0\]
\[ \Rightarrow 2x < 1\]
\[ \Rightarrow x < \frac{1}{2}\]
\[ \Rightarrow x \in \left( 0, \frac{1}{2} \right)\]
\[\text { So,f(x)is decreasing on }\left( 0, \frac{1}{2} \right).\]
\[\text { Since f(x) is increasing on } \left( \frac{1}{2}, 1 \right) \text { and decreasing on }\left( 0, \frac{1}{2} \right),f\left( x \right) \text { is neither increasing nor decreasing on } (0, 1).\]
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