Answer to Question #134471 in Calculus for Tarun Joshua Daniel

Question

(x-1)^2 (x+1)^3
Find the maximum and minimum values

Answer 1

$\mathbf{Let\;f(x)=(x-1)^2(x+1)^3,\;then\;derivative \;of\;f(x)\;is\;given\;by-}$

$\mathbf{\dfrac{d}{dx}f(x)=f'(x)=\dfrac{d}{dx}\big((x-1)^2(x+1)^3\big)}$

$\mathbf{=(x-1)^2\dfrac{d}{dx}\big((x+1)^3\big)+(x+1)^3\dfrac{d}{dx}\big((x-1)^2\big)}$

$\mathbf{=(x-1)^2.3.(x+1)^{3-1}+(x+1)^3.2.(x-1)^{2-1}}$

$\mathbf{=3(x-1)^2(x+1)^2+2(x+1)^3(x-1)}$

$=\mathbf{(x-1)(x+1)^2[3(x-1)+2(x+1)]}$

$\mathbf{=(x-1)(x+1)^2[5x-1]}$

$\textbf{Now, the derivative (slope) is zero when f'(x)=0 ,i.e.,}$

$\textbf{when x=1,\;x=-1,\;x= 1/5}$

$\textbf{We\;know\;that\;-}$

Second Derivative Test-

When a function's slope is zero at x, and the second derivative at x is:

-less than 0, it is a local maximum

-greater than 0, it is a local minimum

-equal to 0, then the test fails (there may be other ways of finding out though)

Here the second derivative of f(x) is-

$\mathbf{f''(x)=4(5x^3+3x^2-3x-1)}$

$\mathbf{At \;x=1\;, f''(1)=4(5+3-3-1)=4*4=16>0}$

$\mathbf{\implies there\;is\;a\;local\;minimum\;at\;x=1\;and\;at\;x=1\;minimum}$

$\mathbf{value\;of\;the\;function\;is-f(1)=(1-1)^2(1+1)^3=0.}$

$\mathbf{At \;x=-1\;, f''(-1)=4(-5+3+3-1)=4*0=0}$

$\mathbf{\implies the\;test\;fails\;at\;x=-1}$

$\mathbf{\implies there\;is\;a\;saddle\;point\;at\;x=-1,i.e.}$

$\mathbf{\implies there\;is\;neither\;a\;maximum\;nor\;a\;minimum\;at\;x=-1.}$

$\mathbf{At \;x=1/5\;, f''(1/5)=4(5(1/5)^3+3(1/5)^2-3(1/5)-1)}$

$\mathbf{=4*(-36/25)=(-144/25)<0}$

$\mathbf{\implies there\;is\;a\;local\;maximum\;at\;x=1/5\;and\;at\;x=1/5\;maximum}$

$\mathbf{value\;of\;the\;function\;is-f(1/5)=((1/5)-1)^2((1/5)+1)^3}$

$\mathbf{=3456/3125}$