Solution (a) The slope of a horizontal tangent line is 0. Since the derivative of $f$ equals the slope of the tangent line, we need to find the numbers $x$ for which $f^\prime ( x) =0$. $$\begin{eqnarray*} f^\prime ( x) &=& 12x^{2}-24x = 12x (x-2) \\ 12x ( x-2 ) &=& 0 & {\color{#0066A7}{\hbox{\(f^{\,\prime} ( x ) =0.\)}}} \\ x &=&0\hbox{ or }x=2 & {\color{#0066A7}{\hbox{Solve.}}} \end{eqnarray*}$$

Figure 23 $f(x) = 4x^3 -12x^2 +2$

At the points $( 0,f( 0) ) =( 0,2)$ and $( 2,f(2) ) =( 2,-14)$, the graph of the function $f( x) =4x^{3}-12x^{2}+2$ has a horizontal tangent line.

(b) Since $f^\prime(x)=12x (x-2)$ and we want to solve the inequalities $f^\prime(x)>0$ and $f^\prime(x) < 0$, we use the zeros of $f^\prime$, 0 and 2, and form a table using the intervals ($-\infty,0$), (0,2), and ($2, \infty)$.

We conclude $f'(x)>0$ on $(-\infty,0) \cup (2,\infty)$ and $f'(x) < 0$ on $(0,2)$,