Question

Let function $f(x)={{(x-1)}^{2}}{{(x+1)}^{3}}.$ Then, which of the following is false?

• A. There exists a point where $f(x)$ has a maximum value
• B. There exists a point where $f(x)$ has a minimum value
• C. There exists a point where $f(x)$ has neither maximum nor minimum value
• D. All of the above

Answer 1

Answer: (D)

All of the above

Answer 2

Given, $f(x)={{(x-1)}^{2}}{{(x+1)}^{3}}$ $f'(x)={{(x-1)}^{2}}3{{(x+1)}^{2}}+2(x-1){{(x+1)}^{3}}$
$=(x-1){{(x+1)}^{2}}[3(x-1)+2(x+1)]$
$=(x-1){{(x+1)}^{2}}[3x-3+2x+2]$
$\Rightarrow$ $f'(x)=(x-1)\,{{(x+1)}^{2}}\,(5x-1)$ ..(i)
For maxima or minima, $f'(x)=0$
$\Rightarrow$ $(x-1){{(x+1)}^{2}}(5x-1)=0$
$\Rightarrow$ $x=-1,\,\,\,1,\frac{1}{5}$
Again, differentiating E (i) . r. t, x. we get $f''(x)=(x-1)(5x-1)\frac{d}{dx}{{(x+1)}^{2}}$ $+{{(x+1)}^{2}}\frac{d}{dx}(x-1)(5x-1)$
$=(5{{x}^{2}}-6x+1)2(x+1)+{{(x+1)}^{2}}(10x-6)$
$=2[5{{x}^{3}}-{{x}^{2}}-5x+1]+(5{{x}^{3}}+7{{x}^{2}}-x-3)\,(2)$
$\Rightarrow$ $f''(x)=2[10{{x}^{3}}+6{{x}^{2}}-6x-2]$ At $x=-1,\,f''(x)=0$
At $x=\frac{1}{5},\,\,f''(x)<0$ and at $x=1,\,\,f''(x)>0$
So, at $x=-1,$
$f(x)$ has neither maximum not minimum value At $x=\frac{1}{5},\,\,f(x)$ has maximum value
At $x=1,\,\,\,f(x)$ has minimum value