Answer to Question #95333 in Real Analysis for Rajni

Question

Identify the intervals in which the function f on R defined by
f(x) = 4x^3-6x^2-72x+15 is both increasing and decreasing ?

Answer 1

To determine intervals where the function strictly increases or decreases, we need to first find the derivative of the function

$f(x) = 4 x^3 - 6 x^2 - 72 x + 15$

$\frac{d}{dx}f(x)=\frac{d}{dx}(4 x^3 - 6 x^2 - 72 x + 15) = 12 x^2 -12x -72=12( x^2 -x -6)=12(x-3)(x+2)$ .

Putting $f'(x)=0.$

$12(x-3)(x+2)=0 \\ (x-3)(x+2)=0$

So $x=3$ and $x=-2$ .

The points divide the real line into 3 disjoint intervals, i.e.

$(-\infty; -2)\bigcup (-2;3)\bigcup (3;\infty)$

$\def\arraystretch{1.5} \begin{array}{c|c|c|c} \text{ interval} & (-\infty;-2) & (-2;3) &(3;\infty) \\ \hline \text{value } x& x<-2 & -2<x<3 & x>3 \\ \hline \text{sign of } f'(x)=12(x-3)(x+2)& (-)(-)=+>0&(-)(+)=-<0&(+)(+)=+>0\\ \hline \text{nature of function } f & \text{ increasing } &\text{ decreasing} & \text{ increasing } \end{array}$

f(x) is increasing in $(-\infty; -2)\bigcup (3;\infty)$

f(x) is decreasing in $(-2;3)$