if ( f(x)=x(x + 1)(x - 2)^2 ), then the graph of ( f ) has inflection points where ( x = ) \na -1 only \nb 2…

Explanation:

Step1: Expand ( F(x) )

First, expand ( F(x)=x(x + 1)(x - 2)^{2} ). [ \begin{align*} F(x)&=x(x + 1)(x^{2}-4x + 4)\ &=(x^{2}+x)(x^{2}-4x + 4)\ &=x^{4}-4x^{3}+4x^{2}+x^{3}-4x^{2}+4x\ &=x^{4}-3x^{3}+4x \end{align*} ]

Step2: Find the first - derivative ( F^{\prime}(x) )

Using the power rule ( (x^{n})^\prime=nx^{n - 1} ), we have ( F^{\prime}(x)=4x^{3}-9x^{2}+4 ).

Step3: Find the second - derivative ( F^{\prime\prime}(x) )

Differentiate ( F^{\prime}(x) ) again. ( F^{\prime\prime}(x)=12x^{2}-18x=6x(2x - 3) ).

Step4: Find the inflection points

Set ( F^{\prime\prime}(x) = 0 ), then ( 6x(2x - 3)=0 ). Solving for ( x ), we get ( x = 0 ) or ( x=\frac{3}{2} ). But wait, we made a mistake above. Let's use the product rule. If ( y = u\cdot v\cdot w) (here (u=x), (v=x + 1), (w=(x - 2)^{2})), then (y^\prime=u^\prime vw+uv^\prime w+uvw^\prime). (u^\prime = 1), (v^\prime=1), (w^\prime = 2(x - 2)) (F^{\prime}(x)=(x + 1)(x - 2)^{2}+x(x - 2)^{2}+2x(x + 1)(x - 2)=(x - 2)[(x + 1)(x - 2)+x(x - 2)+2x(x + 1)]) (=(x - 2)(x^{2}-x - 2+x^{2}-2x+2x^{2}+2x)=(x - 2)(4x^{2}-x - 2)) Now, (F^{\prime\prime}(x)=(x - 2)(8x - 1)+(4x^{2}-x - 2)) (=8x^{2}-x-16x + 2+4x^{2}-x - 2=12x^{2}-18x=6x(2x - 3)) Another way: (F(x)=x(x + 1)(x - 2)^{2}), (F^{\prime}(x)=(x + 1)(x - 2)^{2}+x(x - 2)^{2}+2x(x + 1)(x - 2)) (F^{\prime\prime}(x)=2(x - 2)(x + 1)+(x - 2)^{2}+2(x - 2)x+(x - 2)^{2}+2(x + 1)(x - 2)+2x(x + 1)) (=6(x - 2)(x)+(x - 2)^{2}+2(x + 1)(x - 2)) (=(x - 2)[6x+(x - 2)+2(x + 1)]=(x - 2)(9x))

Set (F^{\prime\prime}(x)=0), then (x = 0) or (x = 2)

Answer:

B. (2) only