The function f(x)=x^2-x√2 is strictly decreasing in the interval :

The correct answer is Option (3) → \( \left(-\infty,\displaystyle\frac{1}{\sqrt{2}}\right] \)

The function is decreasing where its derivative is negative:

    \( \displaystyle f'(x)=2x-\sqrt{2} \)

    \( \displaystyle f'(x)\lt 0\ \Rightarrow\ x\lt\frac{\sqrt{2}}{2} \)

    \( \displaystyle \frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \)

So f is strictly decreasing on \( \left(-\infty,\frac{1}{\sqrt{2}}\right) \). At the single point \( x=\frac{1}{\sqrt{2}} \), \( f'(x)=0 \), but f is still smaller there than at every point to its left, so f is strictly decreasing on \( \left(-\infty,\frac{1}{\sqrt{2}}\right] \).