Question

Let $f\left(x\right)=x^3-3x^2+4x-3$ then find the value of $f\left(\sqrt{2}\right)$.

Answer 1

Given $f\left(x\right)=x^3-3x^2+4x-3$

Now, $f\left(\sqrt{2}\right)=(\sqrt{2}{)}^3-3(\sqrt{2}{)}^2+4\sqrt{2}-3$

$=2\sqrt{2}-6+4\sqrt{2}-3$

$=6\sqrt{2}-9$.