Answer:
We have to determine the graph of the given inequality as:
[tex]x^2+2x+1>0[/tex]
on solving this equation we get:
[tex](x+1)^2>0[/tex]
Since we know that:
[tex](x+1)^2=x^2+1^2+2\times x\times 1\\\\(x+1)^2=x^2+1+2x[/tex]
Now we know that when [tex]x=-1[/tex] the value of:
[tex](x+1)^2=0[/tex]
Also for the value other than -1 the function being a quadratic function will always give a non-zero positive value.
Hence, range of the function in intervals could be written as:
(-∞,-1)∪(-1,∞).
Hence, the graph of the function will be the whole of the number line with a open circle on -1.
