Question

if ( f(x)=x(x + 1)(x - 2)^2 ), then the graph of ( f ) has inflection points where ( x = )
a -1 only
b 2 only
c -1 and 0 only
d -1 and 2 only
e -1, 0, and 2 only

Explanation:

Step1: Expand \( F(x) \)

First, expand \( F(x)=x(x + 1)(x - 2)^{2} \).

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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