Question

If $p\left(y\right)=3y^3-4y+\sqrt{11}$, find the value of p(y) at $y=2$.

Answer 1

Given that, $p\left(y\right)=3y^3-4y+\sqrt{11}$

Substituting $y=2$ in $p\left(y\right)$, we get:

$p\left(y\right)=3{\left(2\right)}^3-4\times 2+\sqrt{11}$

$\Rightarrow p\left(y\right)=24-8+\sqrt{11}$

$\therefore p\left(y\right)=16+\sqrt{11}$

Hence, the value of $p\left(y\right)$ at $y=2$ is $16+\sqrt{11}$.