The function f(x)=2x^3-30x^2+96x-3 has two critical numbers. The smaller one is x=□ and the larger one is x=□

2 Find the derivative of the function using the power rule for derivatives, which states that the derivative of $f(x)=ax^{n}$ is $f'(x)=nax^{n-1}$

3 Apply the power rule to the given function $$f(x)=2x^{3}-30x^{2}+96x-3$$ to get the derivative $f'(x)=6x^{2}-60x+96$

4 Set the derivative equal to zero and solve for x to find the critical numbers. The equation $6x^{2}-60x+96=0$ can be simplified by dividing each term by 6 to get $x^{2}-10x+16=0$

5 Factor the quadratic equation to find the values of x that satisfy the equation. The factors are (x-2)(x-8)=0 , which gives us the solutions x=2 and x=8

6 Conclude that the smaller critical number is x=2 and the larger critical number is x=8