Mary Reyes

2021-12-31

Find range of the function
$f\left(x\right)=3|\mathrm{sin}x|-4|\mathrm{cos}x|$
I tried to do by using the trigonometric identities

So
$f\left(x\right)=3\sqrt{\frac{1-\mathrm{cos}2x}{2}}-4\sqrt{\frac{1+\mathrm{cos}2x}{2}}$
but don't know how to proceed from here

Kayla Kline

and
$-4\le f\left(x\right)\le 3$,
for all $x\in \mathbb{R}$. Intermediate value theorem gives $f\left(\mathbb{R}\right)=\left[-4,3\right]$.

Mollie Nash

Let
$f\left(x\right)=3|\mathrm{sin}x|-4|\mathrm{cos}x|$
It's easy to show f(x) is periodic with $T=\pi$ and f(x) is even. So it's enough to consider $0\le x\le \frac{\pi }{2}$
If $0\le x\le \frac{\pi }{2}$ then
$0\le \mathrm{sin}x\le 1⇒0\le |\mathrm{sin}x|\le 1$
And
$0\le \mathrm{cos}x\le 1⇒0\le |\mathrm{cos}x|\le 1$
Therefore we have $-4\le f\left(x\right)\le 3$ but this could be only an upper bound not the actual range.
If $0\le x\le \frac{\pi }{2}$ then
$\left(x\right)=3\mathrm{sin}x-4\mathrm{cos}x⇒f\prime \left(x\right)=3\mathrm{cos}x+4\mathrm{sin}x>0⇒$
$f\left(0\right)\le f\left(x\right)\le f\left(\frac{\pi }{2}\right)⇒-4\le f\left(x\right)\le 3$
Confirming the first result. Actually because f(x) is an increasing function in that interval, the results are the same. So we can conclude that $-4\le f\left(x\right)\le 3$ and WA verifies that.

Vasquez

$\begin{array}{}f\left(x\right)=3|\mathrm{sin}x|-4|\mathrm{cos}x|\\ -4\le -4|\mathrm{cos}x|\le f\left(x\right)\le \\ 3|\mathrm{sin}x|\le 3\\ \text{By inspection:}\\ f\left(\frac{\pi }{2}\right)=3;f\left(0\right)=-4\\ {\text{Range}}_{f}=\left[-4,3\right]\end{array}$