91Ó°ÊÓ

The given function is$g\left(x\right)=6x^2-18x.$

The given function is positive for the values of xwhen g(x) > 0,and negative for the values of xwhen g(x) < 0.

Now, to find the intervals put g(x) = 0,

$$\begin{gathered} g\left(x\right)=6x^2-18x \\ 0=6x\left(x-3\right) \\ x=0\mathrm{and}x=3 \end{gathered}$$

Sign chart of the function is

Thus, the function is positive in the intervals $(−\mathrm{∞},0)\text{and}(3,\mathrm{∞}),$and negative in the intervals$\left(0,3\right).$

The given function is positive for the values of xwhen g(x) > 0,and negative for the values of xwhen g(x) < 0.

Now, to find the intervals put g(x) = 0,

$$\begin{gathered} g\left(x\right)=2-x-2x^2+x^3 \\ 0=2-x-2x^2+x^3 \\ 0=\left(x-1\right)\left(x+1\right)\left(x-2\right) \\ x=1,x=-1,\mathrm{and}x=2 \end{gathered}$$

Sign chart of the function is

Thus, the function is positive in the intervals $(−1,1)\text{and}(2,\mathrm{∞}),$and negative in the intervals$(−\mathrm{∞},−1)\text{and}(1,2).$

The given function is positive for the values of xwhen g(x) > 0,and negative for the values of xwhen g(x) < 0.

Since the function $e^x\text{and}(x−1{)}^4$ are always on the real line, thus g(x) > 0,

$$\begin{gathered} \left(x+2\right)>0 \\ x>-2 \end{gathered}$$

And g(x) < 0,

$$\begin{gathered} \left(x+2\right)<0 \\ x<-2 \end{gathered}$$

Sign chart of the function is

Thus, the function is positive in the interval $(2,\mathrm{∞}),$and negative in the interval$(−\mathrm{∞},−2).$

The given function is positive for the values of xwhen g(x) > 0,and negative for the values of xwhen g(x) < 0.

Since the function ${\sin }^{2}x$ are always on the real line, g(x) > 0 andg(x) < 0,

$$\begin{gathered} 3x^2-5x-2>0 \\ \left(x-1\right)\left(3x-2\right)>0 \\ \mathrm{And} \\ 3{\mathrm{x}}^2-5\mathrm{x}-2<0 \\ \left(\mathrm{x}-1\right)\left(3\mathrm{x}-2\right)<0 \end{gathered}$$

To find the intervals let's put g(x) = 0,

$$\begin{gathered} \left(x-1\right)\left(3x-2\right)=0 \\ x=1\mathrm{and}x=\frac{2}{3} \end{gathered}$$

Sign chart of the function is

Thus, the function is positive in the intervals $(−\mathrm{∞},1)\text{and}(\frac{2}{3},\mathrm{∞})$and negative in the interval$\left(1,\frac{2}{3}\right).$