Question

A function f is defined on [-1,1] as
f(x)={-x+1/2 ; -1≤x {x+1/2. ; 0≤x≤1
And if g(x) = |f(x)|+f(|x|) then find
1.) Number of points of discontinuity of g(x).
2.) Maxima & Minima of g(x).
3.) Domain and range of g(x).

Answer 1

Dear Student,


$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{ll}-x+\frac{1}{2}& -1\le x\\ x+\frac{1}{2}& 0\le x\le 1\end{array}\right. \\ nowg\left(x\right)=\left|f\left(x\right)\right|+f\left(\left|x\right|\right) \\ nowf\left(\left|x\right|\right)willalwayscontainsx+\frac{1}{2}as\left|x\right|\ge 0 \\ Henceg\left(x\right)=\left\{\begin{array}{ll}-x+\frac{1}{2}+x+\frac{1}{2}& -1\le x\\ x+\frac{1}{2}+x+\frac{1}{2}& 0\le x\le 1\end{array}\right. \\ \Rightarrow g\left(x\right)=\left\{\begin{array}{ll}1& -1\le x\\ 2x+1& 0\le x\le 1\end{array}\right. \\ i)Numberofpointsofdiscontinuityofg\left(x\right) \\ checkingcontinuityatx=0 \\ atx=0,g\left(x\right)=1 \\ atx=-1,g\left(x\right)=1 \\ atx=1,g\left(x\right)=2\times 1+1=3 \\ hencebetween\left[-1,1\right]thereisnopointwhereg\left(x\right)isnotdefined. \\ Hencenumberofpointsofdisconituity=0 \\ ii)Minimumvalueofg\left(x\right)=1whenx=0andx=-1 \\ maximumvalueofg\left(x\right)=2\times 1+1=3atx=1 \\ iii)domainofg\left(x\right)is\left[-1,1\right]asitisdefinedateachpointinx\in \left[-1,1\right] \\ rangeisthemaximumandminimumvaluesofg\left(x\right) \\ hencerange=\left[1,3\right] \end{gathered}$$
Regards,