The number of extremum point(s) for f(x)=3|x|+|x+1|−||x+1|−3|x|| is
Open in App
Solution
Given : f(x)=3|x|+|x+1|−||x+1|−3|x||
We know, f(x)+g(x)−|f(x)−g(x)|=min{2f(x),2g(x)}
Here, f(x)=|x+1|,g(x)=3|x|
where: 2x+2=0 gives x=−1, 6x=0 gives x=0, 2x+2=6x gives x=12 and 2x+2=−6x gives x=−14
Now, plotting the graph of y=min{|2x+2|,6|x|}: Clearly, it has 3 extremum points.