Question

find the critical numbers for ( h(x)=3x^{3}-\frac{3}{2}x^{2}-2x ).

Explanation:

Step1: Find the derivative of the function

The derivative of \(h(x)=3x^{3}-\frac{3}{2}x^{2}-2x\) is \(h'(x)=9x^{2}-3x - 2\) (using the power rule \((x^{n})'=nx^{n - 1}\)).

Step2: Set the derivative equal to zero

Set \(h'(x)=0\), so \(9x^{2}-3x - 2=0\).
For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 9\), \(b=-3\), \(c=-2\)), the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) is used.
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-3)^{2}-4\times9\times(-2)=9 + 72=81\).
Then \(x=\frac{-(-3)\pm\sqrt{81}}{2\times9}=\frac{3\pm9}{18}\).

Step3: Solve for \(x\)

For the plus - case: \(x=\frac{3 + 9}{18}=\frac{12}{18}=\frac{2}{3}\).
For the minus - case: \(x=\frac{3-9}{18}=\frac{-6}{18}=-\frac{1}{3}\).

Answer:

The critical numbers are \(x =-\frac{1}{3}\) and \(x=\frac{2}{3}\).