Question

Show that the function x² − x + 1 is neither increasing nor decreasing on (0, 1).

Answer 1

$$\begin{gathered} f\left(x\right)=x^2-x+1 \\ f\text{'}\left(x\right)=2x-1 \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow 2x-1>0 \\ \Rightarrow 2x>1 \\ \Rightarrow x>\frac{1}{2} \\ \Rightarrow x\in \left(\frac{1}{2},1\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(\frac{1}{2},1\right). \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow 2x-1<0 \\ \Rightarrow 2x<1 \\ \Rightarrow x<\frac{1}{2} \\ \Rightarrow x\in \left(0,\frac{1}{2}\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(0,\frac{1}{2}\right)\text{.} \\ \text{Since}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on}\left(\frac{1}{2},1\right)\text{and decreasing on}\left(0,\frac{1}{2}\right)\text{,}f\left(x\right)\text{is neither increasing nor decreasing on (0, 1).} \end{gathered}$$