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Calculate the range of the function $f$ with $f(x) = x^2 - 2x$, $x\in\Bbb{R}$. My book has solved solutions but I don't get what is done:

$$f(x) = x^2 - 2x + (1^2) - (1^2)= (x-1)^2 -1$$

edit: sorry for wasting all of the people who've answered time, I didnt realize that I had to complete the square.... Wish I could delete it but I cant

thezaynthe
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  • The author completed the square in order to show that the graph of $f(x)$ is a parabola with vertex $(1, -1)$ that opens upwards. – N. F. Taussig Dec 24 '15 at 11:03
  • As you know, $(x - k)^2 = x^2 - 2kx + k^2$. Since $f(x) = x^2 - 2x = x^2 - 2(1)x$, we can obtain a perfect square by adding $1^2$. To balance the equation, we must also subtract $1^2$. – N. F. Taussig Dec 24 '15 at 11:33

3 Answers3

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From the form $f(x)=(x-1)^2-1$ you can see immediately two things:

  1. Since the term $(x-1)^2$ is a square it is always non-negative, or in simpler words $\ge 0$. So, its lowest possible values is $0$ which can be indeed attained when $x=1$. So, $f(x)=\text{ positive or zero } -1 \ge 0-1=-1$. So $f(x)$ takes values greater than $-1$. So this the lower bound for the range of $f$.
  2. Is there also an upper bound for the values of $f$, or can $f(x)$ take all values that are greater than $-1$? To answer this, you should ask if you can increase the first term $(x-1)^2$ (that depends on $x$) as much as you want. The second term $-1$ is now irrelevant. Indeed $(x-1)^2$ gets bigger constantly as $x$ increases and in fact without an end.

This allows you to conclude that $$\text{Range}(f)=[-1, +\infty)$$ and that is why they brought $f$ to this form.

Jimmy R.
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For any $y\in\Bbb{R}$, $y^2\ge0$. So $y^2-1\ge-1$.

Substitute $y=x-1$. What does it imply?

Mythomorphic
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Hint:

i suppose that you are convinced that $f(x)=x^2-2x=(x-1)^2-1$. Now write this as: $f(x)=-1+(x-1)^2$ and note that the square is always not negative, and it is null for $x=1$.

So what you can say about $f(x)$?

Emilio Novati
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