Question

The function $f\left(x\right)=2-3x+3x^2-x^3,x\epsilon R$ is

• A. neither increasing nor decreasing
• B. increasing
• C. decreasing
• D. none of these

Answer 1

Answer: (A)

neither increasing nor decreasing

Consider given the function,

$f\left(x\right)=2-3x+3x^2-x^{3}x\in R$

$f^{{}^{\prime }}\left(x\right)=-3+6x-3x^2$

For increasing and decreasing,

$f^{{}^{\prime }}\left(x\right)=0$

$-3+6x-3x^2=0$

$x^2-2x+1=0$

$x^2-x-x+1=0$

$x(x-1)-1(x-1)=0$

${(x-1)}^2=0$

$x=1,1$

Hence, x=1 divided the function into two part $(-\infty ,1)$ and$(1,\infty )$

For, part $(-\infty ,1)$ function will be decrease,

And $(1,\infty )$ function will be increase.

Hence, function neither increase nor decrease.