I do not get this question at all. I need to prove the an equation has a minimum. Quadratics involved.

Question

Prove that $f(x)= (x-a)^2+(x-b)^2$ has a minimum when $x= \frac{a+b}{2}$. (Prove not verify) I do not get this question whatsoever, please help me.

Answer 1

$f'(x)=2(x-a)+2(x-b)=4x-2(a+b)$ and $f''(x)=4$

Theorem: If $f'(c)=0$ and $f''(c)> 0$ then $c$ is a strict local minimum.

Now check this for $c=\frac{a+b}{2}$. Well, indeed, $f'(\frac{a+b}{2})=0$. CHECK! And indeed $f''(\frac{a+b}{2})=4>0$ CHECK! Thus, by the theorem, we may conclude that $c=\frac{a+b}{2}$ is indeed a strict local minimum (meaning you can find a neighborhood around $c$ such that $f(c)$ is indeed the minimum value that $f(x)$ takes in that neighborhood.

But we can say more! Since the $f'(x)\neq 0$ anywhere else on the real line, and the second derivative is strictly positive everywhere, we can safely conclude that $c=\frac{a+b}{2}$ is a global minimum of $f(x)$ on the real line. In other words, we found the smallest value that $f(x)$ takes on the whole real line (the domain), and that is at $c$.

Answer 2

There are a few ways you can attack this problem.

Method 1

Expand the expression, collect like terms, and get the formula into the form $f(x)=Ax^2 + Bx + C$ (where $A, B$, and $C$ are some constants, possibly involving combinations of the $a$ and $b$ in the original problem.) Then invoke two facts:

1. If $A$ is positive, then the graph is an upwards-pointing parabola; and
2. The vertex of that parabola occurs at $x=-B/2A$.

Both of these are facts from first-year Algebra; if you are allowed to use them, then they together tell you where the minimum of the function occurs.

Method 2

First, plug in the given value $x=\frac{a+b}{2}$ into the formula, and compute what $f(x)$ is actually equal to for that given value. (The value will involve $a$ and $b$.)

Next, suppose $x \ne \frac{a+b}{2}$. Then $x=\frac{a+b}{2} + k$ for some $k\ne 0$. Plug that into the formula for $f(x)$ and compute what the value of $f(x)$ is for all values other than the given one. (The value will involve $a$, $b$ and $k$.)

Now compare those two results and explain why the second one must always be larger than the first one as long as $k \ne 0$.

Method 3

Since the function is quadratic, and the coefficient of $x^2$ in both terms is positive, the function is an upwards-pointing parabola; therefore it has to have a minimum somewhere. Now, notice that the function is "symmetric in $a$ and $b$". This means that if you swap the values of $a$ and $b$ in the function, the function itself does not change. Graphically, that is equivalent to reflecting the parabola across a mirror halfway between $a$ and $b$. The fact that the function is symmetric across that mirror tells us that the vertex of the parabola must be on that mirror -- which leads directly to the desired conclusion.

Answer 3

Without calculus:

Following user254665’s suggestion, expand, collect terms, and complete the square $$\begin{align} (x-a)^2+(x-b)^2 &= x^2-2ax+a^2 + x^2-2bx+b^2 \\ &= 2x^2-2(a+b)x+(a^2+b^2) \\ &= 2\left(x^2-(a+b)x+\left(\frac{a+b}2\right)^2\right)-2\left(\frac{a+b}2\right)^2+(a^2+b^2) \\ &= 2\left(x-\frac{a+b}2\right)^2+\frac{(a-b)^2}2. \end{align}$$ From here, you should be able to figure out where the non-negative term on the left has its minimum.

Answer 4

You probably know that a quadratic equation

$$y=Ax^2+Bx+C$$

can be represented by a parabola on the xy plane. You probably also know that if $A>0$ then it opens upwards and if $A<0$ then it opens downwards. Okay, so keeping that in mind, a parabola that opens upwards must have a minimum at the vertex.

Expanding your equation, $$\begin{align} f(x)&= (x-a)^2+(x-b)^2 \\ &=x^2 -2ax+a^2 + x^2 -2bx+b^2 \\ &=2x^2 -2(a+b)x+(a^2+b^2) \end{align} $$

we see that $A=2>0$, so it opens upwards and has a minimum.

Next you need to actually find the minimum. You may have learned vertex form (sometimes this goes by a different name):

$$y=c(x-h)^2+k,\quad\quad\text{where $(h,k)$ is the vertex}$$

If you can rewrite your given equation in vertex form, you can read off the vertex $(h,k)$, which is also your minimum. Good luck from here!

Hint:

The process that I'm suggesting is called completing the square.