Answer:
at x = [tex]\sqrt{2}[/tex], [tex]-3x^{2} + \sqrt{2}x + \frac{1}{2}[/tex] = -3.5
Step-by-step explanation:
The given equation is [tex]-3x^{2} + \sqrt{2}x + \frac{1}{2}[/tex]
Now, substitute the value of x = [tex]\sqrt{2}[/tex] in the above equation.
The given equation simplifies to,
[tex]-3(\sqrt{2})^{2} + (\sqrt{2})(\sqrt{2}) + \frac{1}{2}[/tex]
now, [tex]\sqrt{2} . \sqrt{2} = 2, \textrm{and} (\sqrt{2} )^{2} = 2[/tex]
So, the equation become
-3(2) + 2 + (1/2) = [tex]-6 + 2 + \frac{1}{2} = -4 + \frac{1}{2} = \frac{-8 + 1}{2} = \frac{-7}{2}[/tex]
= -3.5
So, at x = [tex]\sqrt{2}[/tex], [tex]-3x^{2} + \sqrt{2}x + \frac{1}{2}[/tex] = -3.5
