What is the least value of the function $y= (x-2) (x-4)^2 (x-6) + 6$? What is the least value of function:
$$y= (x-2)  (x-4)^2  (x-6) + 6$$
For real values of $x$ ?
Does $\frac{dy}{dx} = 0$, give the value of $x$ which will give least value of $y$?
Thanks in advance.
 A: $$\frac{dy}{dx} = (x-4)^2 (x-6)+(x-4)^2 (x-2)+2(x-2)(x-4)(x-6) = 0 $$
$$ \Rightarrow (x-4) \left[ (x-4)(2x-8)+2(x-2)(x-6) \right] = 0$$
$$ \Rightarrow (x-4) \left[ 2x^2-16x+32+2x^2-16x+24 \right] = 0$$
$$ \Rightarrow (x-4) \left[ 4x^2-32x+56 \right] = 0$$
$$ \Rightarrow 4(x-4) \left[ (x-4+\sqrt{2})(x-4-\sqrt{2}) \right] = 0$$
There is a maxima or minima at $x=4, \hspace{5pt}x=(4-\sqrt{2}),\hspace{5pt} x=(4+\sqrt{2})$ 
If you take the second derivative and find where the second derivative is positive, that is where the function value has a minima.
Second derivative is 
$$8(x-4)^2+4(x-4+\sqrt{2})(x-4-\sqrt{2})$$
And at $x=4\pm\sqrt{2}$ for instance the second derivative is positive and therefore at it is minimum at those points.
A: An alternative solution using AM-GM inequality instead of derivatives.

If you use substitution $t=x-4$, then you want to minimize $t^2(t-2)(t+2)+6$, which is equivalent to minimizing
$$g(t)=t^2(t-2)(t+2)=t^2(t^2-4).$$
Obviously $g(t)\ge 0$ whenever $t^2\ge 4$. We can use another substitution $s=t^2$ and ask for minimum of $s(s-4)$ for $s\in\langle 0,4 \rangle$. 
If we change the sign, this is the same thing as looking for the maximum of 
$$h(s)=s(4-s)$$
for $s\in\langle 0,4 \rangle$. 
Now we can use AM-GM inequality to see that
$$h(s)=s(4-s) \le \left(\frac{s+(4-s)}2\right)^2 = 2^2=4,$$
where the equality is attained only for $s=4-s$, i.e. $s=2$.
This leads to $t=\pm\sqrt2$ and $x=4\pm\sqrt2$. The minimal value is 2. (Obtained as $6-4=2$, since $4$ is the maximal possible value of $h(s)$.)

You can check this at wolframalpha: minimize (x-2) * (x-4)^2 * (x-6) + 6. 
A: Since you are asked to find the least value of 
$$\tag{1}
f(x)=(x-2)(x-4)^2(x-6)+6
$$
over the reals, you are looking for the  global minimum value, if there is one.

Your function is continuous, so the following general procedure is applicable:
Given a continuous function $g$ defined on a nondegenerate interval $I$, to determine if $g$ has a global minimum value on $I$ and to find its value when it exists:


*

*Examine the "endpoint behaviour" of $g$. For example:


*

*If $I=[a,b]$, evaluate $g(a)$ and $g(b)$.

*If $I=(a,b)$, calculate $\lim\limits_{x\rightarrow a^+} g(x)$ and    $\lim\limits_{x\rightarrow b^-} g(x)$.

*If $I= \Bbb R$, calculate $\lim\limits_{x\rightarrow\infty} g(x)$ and    $\lim\limits_{x\rightarrow-\infty} g(x)$.
There are other cases to consider, but I hope that it's obvious what
to do in those cases.
The quantities found here will be used later, so keep them in mind.

*Find all critical points of $g$ in the interval $I$. These are
points $x\in I$ where $g'(x)=0$ or where $g'(x)$ does not exist. These points and any endpoints of $I$ contained in $I$ are the only points at which a global minimum value can occur.

*Evaluate the function $g$ at each point found in step 2..

*Use the information found in steps 1. and 3. to determine if $g$
has a global  minimum value and what that value is when it exists.
That, is compare the values found in steps 1. and 3.. There are
several cases to consider depending on the form of $I$.  For
example:


*

*If $I=[a,b]$, then $g$ will have a global minimum value. The global    minimum will be the smallest of the quantities found in
steps 1. and    3..

*If $I=(a,b)$, then $g$ will have a global minimum value if and only if the smallest value found in step 3. is less than or equal to
both of the limit values found in step 1.. In this case, the global
minimum value of $g$ is the smallest value found in step 3.. Note
here that since $I$ does not include its endpoints,  the endpoints of $I$ are not  candidates for places where the global minimum of $g$, if it exists, can occur.

*If $I=\Bbb R$, then $g$ will have a global minimum value if and only    if the smallest value found in step 3. is less than or equal
to both    of limit values found in step 1.. In this case, the global
minimum value of $g$ is the smallest value found in step 3..

Back to our problem:
In our case, the implied domain of $f$ is all of $\Bbb R$. 
Step 1:
Note that the function $f$ is a polynomial of degree 4 with positive leading coefficient. This implies that 
$$\lim\limits_{x\rightarrow\infty}f(x)=\infty\quad\text{and}\quad
 \lim\limits_{x\rightarrow-\infty}f(x)=\infty.$$
Looking ahead a bit, we can conclude that $f$ does have a global minimum value and that value will be  the smallest value found in step 3..
Step 2:
We first find
$$\tag{2}\eqalign{
f'(x)&=(x-4)^2(x-6)+2(x-2)(x-4)(x-6)+(x-2) (x-4)^2\cr
&=4(x-4)(x^2-8x+14 ).
}$$
The derivative $f'$ exists everywhere. We are left with the task of solving  the equation $f'(x)=0$. This is easy to do looking at $(2)$ and using the quadratic formula: $f'(x)=0$ if and only if
$$
x=4,\quad x={8\pm\sqrt{64-56}\over 2}=4\pm\sqrt2.
$$
Step 3: 
We evaluate $f$ at each of the three points found above: 
$$\tag{3}
\eqalign{
f(4)&= 6\cr
f(4+\sqrt2)&=(2+\sqrt2)\cdot2\cdot( -2+\sqrt2)+6=2\cr
f(4-\sqrt2)&=(2-\sqrt2)\cdot2\cdot(-2-\sqrt2)+6=2.\cr
}
$$
Step 4:
From step 1., $f(x)$ tends to infinity as $x\rightarrow\infty$ or as  $x\rightarrow-\infty$.
This means that $f$ does have a global minimum value and that that value must be the smallest value found in step 3..
So, from $(3)$, we see that the minumum value of $f$ is $2$. This minimum value is achieved  at two places: $x=4+\sqrt2$ and at $x=4-\sqrt 2$.
A: Note that $(x-2)(x-6)=x^2-8x+12$ and $(x-4)^2=x^2-8x+16$. This suggests the symmetrizing substitution $w=x^2-8x+14$. Thus 
$$y=(w-2)(w+2)+6=w^2+2.$$ 
We want to minimize the absolute value of $w$. But $w=(x-4)^2-2$. So $w$ has minimum absolute value $0$, reached when $(x-4)^2=2$, that is, when $x=4\pm\sqrt{2}$.  The minimum value of $y$ is $2$.  
Remark: The curve $y=(x-2)(x-4)^2(x-6)+6$ is very special, with a beautiful double symmetry. It seems likely that the method used above was the intended one. The same idea works for $(x-a)(x-b)(x-c)(x-d)+k$, where $a+d=b+c$. 
A: $$y= (x-2)  (x-4)^2  (x-6) + 6$$
$let , t = x-4$
$$y= t^2(t-2)(t+2)+ 6$$
$$y= t^4-4t^2+6 $$
$$y=(t^2-2)^2+2$$
$$y(min)=2$$ 
attained when $t^2=2$
$(x-4)^2=2$ which gives, $ x=4-\sqrt2$ and $ x=4+\sqrt2$
