Question

j(x)=−6x^3+2x / 5x^2−3. which of the following is true?j(2)>j(0)j times 2 is greater than j times 0j(0)>j(−2)j times 0 is greater than j times negative 2j(−2)>j(0)j times negative 2 is greater than j times 0j(2)>j(−2)j times 2 is greater than j times negative 2

Answer 1

The given function is,

$j(x)=\frac{-6x^3+2x}{5x^2-3}\text{ ---(1)}$

To find j(0), substitute x=0 in expression (1).

$\begin{gathered} j(0)=\frac{-6\cdot0^3+2\cdot0}{5\cdot0^2-3}\text{ } \\ j(0)=0 \end{gathered}$

So, j(0)=0.

To find j(2), substitute x=2 in expression (1).

$\begin{gathered} j(2)=\frac{-6\cdot2^3+2\cdot2}{5\cdot2^2-3}\text{ } \\ j(2)=\frac{-6\cdot8+4}{5\cdot4-3} \\ j(2)=\frac{-48+4}{20-3} \\ j(2)=\frac{-44}{17} \\ j(2)=-2\frac{10}{17} \end{gathered}$

Therefore, j(2)=-2 10/17.

To find j(-2), substitute x=-2 in expression (1).

$\begin{gathered} j(-2)=\frac{-6\cdot(-2)^3+2\cdot(-2)}{5\cdot(-2)^2-3}\text{ ---(1)} \\ j(-2)=\frac{-6\cdot(-8)-4}{5\cdot4-3} \\ j(-2)=\frac{48-4}{20-3} \\ j(-2)=\frac{44}{17} \\ j(-2)=2\frac{10}{17} \end{gathered}$

Therefore, j(-2)=2 10/17.

So, the obtained values are,

$\begin{gathered} j(0)=0 \\ j(2)=-2\frac{10}{17} \\ j(-2)=2\frac{10}{17} \end{gathered}$

Comparing the values of j(0), j(2) and j(-2), we get

j(2)

Therefore, the statement j(-2)> j(0) is correct.