Rewrite the following into a piecewise function without any absolute values:
y = [tex]\mid x-3\mid+\mid5+x\mid[\math]
i don't know if i did latex right, so here it is in regular text
y=(absolute value x-3) + (absolute value 5+x)
Rewrite the following into a piecewise function without any absolute values:
y = [tex]\mid x-3\mid+\mid5+x\mid[\math]
i don't know if i did latex right, so here it is in regular text
y=(absolute value x-3) + (absolute value 5+x)
Are you required to draw the graph as well? I thought you were only asked to rewrite the function without using absolute values.
Write the function in the form $\displaystyle y=\begin{cases}\ldots & x\le-5\\\\
\ldots & -5\le x\le 3\\\\
\ldots & x\ge3\end{cases}$.
Hello, Kingreaper!
Solve: .$\displaystyle \sqrt{2x+1} + \sqrt{2x+6} \:=\:5$
Isolate a radical: .$\displaystyle \sqrt{2x+1} \;=\;5 - \sqrt{2x+6}$
Square both sides: .$\displaystyle \left(\sqrt{2x+1}\right)^2 \;=\;\left(5 - \sqrt{2x+6}\right)^2$
. . . . . . . . . . . . . . . . . $\displaystyle 2x + 1 \;=\;25 - 10\sqrt{2x+6} + 2x + 6$
. . . . . . . . . . . . . . .$\displaystyle 10\sqrt{2x+6} \:=\:30 $
. . . . . . . . . . . . . . . . $\displaystyle \sqrt{2x+6} \:=\:3$
Square both sides: .$\displaystyle 2x + 6 \:=\:9\quad\Rightarrow\quad 2x \:=\:3 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{3}{2}}$