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)=(-6x^3+2x)/(5x^2-3)\text{ ---(1)}`

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

`\begin{gathered} j(0)=(-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)=(-6\cdot2^3+2\cdot2)/(5\cdot2^2-3)\text{ } \\ j(2)=(-6\cdot8+4)/(5\cdot4-3) \\ j(2)=(-48+4)/(20-3) \\ j(2)=(-44)/(17) \\ j(2)=-2(10)/(17) \end{gathered}`

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

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

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

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

So, the obtained values are,

`\begin{gathered} j(0)=0 \\ j(2)=-2(10)/(17) \\ j(-2)=2(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.