Let $f(x)=x^2-8x+4$, where $x\in[4,\infty)$. Show that $f$ is one-to-one.

Question

I would like to know if this is ok.

Let $f(x)=x^2-8x+4$ with $x\in[4,\infty)$. Show that $f$ is one-to-one.

If,

$$f(a)=f(b)$$ $$a^2-8a+4=b^2-8b+4$$ $$a^2-8a=b^2-8b+4-4$$ $$a^2-8a=b^2-8b$$ $$a(a-8)=b(b-8)$$

Therefore, $$a=b$$

Therefore,$f$ is one-to-one.

Thanks for your time.

Answer 1

$$a(a-8)=b(b-8)$$

Here, it is not sufficient to conclude $a=b$, for example, $a=0, b=8$, so you need to use the domain $[4, \infty)$ to exclude the cases such as $a=0, b=8$.

$$a^2-8a=b^2-8b$$

Go back to this step, instead, you can factor it in this way,

$$\begin{align}a^2-b^2&=8a-8b\\ \\ (a+b)(a-b)&=8(a-b)\\ \\ (a+b-8)(a-b)&=0\end{align}$$

Can you proceed from here?

Answer 2

You are on the right track, but you cannot conclude the desired result as you did.

I would slightly change the last step to obtain: \begin{align*} a^{2} - 8a = b^{2} - 8b & \Longleftrightarrow a^{2} - b^{2} = 8(a - b)\\\\ & \Longleftrightarrow (a - b)(a + b) = 8(a - b)\\\\ & \Longleftrightarrow (a - b)(a + b - 8) = 0 \end{align*}

Can you take it from here?

Answer 3

Note $f(2)=f(6)=-8$. But you're on the right track.

$f(x)=x^2-8x+4$

$a^2-8a+4=b^2-8b+4$

$(a^2-b^2)-8(a-b)=0$

$(a-b)(a+b-8)=0$

So either $a=b$ or $a+b=8$.

Given the domain, either $a=b=4$ or $a+b>8$, so we must have $a=b$.

Alternatively $f(x)=x^2-8x+4=(x-4)^2-12$. So $f(x)$ has a minimum value of $-12$ and increases monotonically from there. $x>y \iff f(x)>f(y) $

Suppose $f(x)=f(y)$ and $x>y$. By the monotonicity, we must have $f(x)>f(y)$, thus the assumption that $x>y$ leads to a contradiction. The same argument applies if we have $f(x)=f(y)$ and $x<y$. Given two quantities $x$ and $y$, either $x=y$, $x<y$ or $x>y$. The two inequalities have been ruled out leaving only equality, so a monotonic function is one to one.

Answer 4

Alternative approach:

$\underline{\text{Intermediate Result}}$
IR-1
For $~y \geq 0,~$ the function $~g(y) = y^2~$ is strictly increasing on $~[0,\infty).$
Proof
Assume that $~0 \leq y_1 < y_2.$
Then $~y_2^2 - y_1^2 = (y_2 - y_1) \times (y_2 + y_1) > 0.$
Thus, $~y_2 > y_1 \implies y_2^2 > y_1^2.$

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It is sufficient to show that
$~f(x) = x^2 - 8x + 4 ~$ is strictly increasing on $[4,\infty).$

$$f(x) = (x - 4)^2 - 12.$$

Let $~y = (x-4) \implies \{ ~[4 \leq x] \iff [0 \leq y] ~\}.$

Then, setting $~g(y) = y^2,~$ you have that
$f(x) = g(y) - 12.~$

By IR-1, $~g(y)~$ is strictly increasing.
Therefore, so is $~f(x) = g(y) - 12.$