Range of the function $f(x) = |x-3|+|x+5|$

Question

As the title suggests, we have to find the range of $f(x) = |x-3|+|x+5|$

I know how to find the range when we have a function in which only one modulus function is involved like $|x\pm a|$ or $k\pm |x\pm a|$, but I don't know how to solve this given question.

Being a newcomer to modulus functions, I asked my teacher this question. He suggested me to find the range by plotting graph of the function. I solved it using the method suggested by my teacher but I am wondering if there's any more feasiable method to find the range of this function.

Any hint would be enough, thanks a ton in advance!

Answer 1

According to the absolute value properties, we have \begin{align*} |x - 3| + |x + 5| = |3 - x| + |x + 5| \geq 3 - x + x + 5 = 8 \end{align*}

Answer 2

We can also do as follows:

Let,

$$y=|x-3|+|x+5|$$

Then, using the substitutions:

$$x+5=a,~x-3=b$$

We have,

$$\begin{align}&\begin{cases}y=|a|+|b|, \thinspace y≥0\\ a-b=8 \end{cases}\\\\ \implies &\begin{cases}a^2+b^2+2|ab|=y^2\\a^2+b^2-2ab=8^2\end{cases}\\\\ \implies &y^2-8^2=2\left(|ab|+ab\right)≥0\\\\ \implies &y≥8.\end{align}$$

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Also we see that,

• If $x≥3$ or $x≤-5$ then $y≥8$.

• If $-5≤x≤3$, then $y=3-x+x+5=8$.

This means, $y\in[8,+\infty)$.