Given thatf(x)=⎩⎨⎧x2+3x+2x+2,−1,0,if x∈R−{−1,−2}if x=−2if x=−1.Now, we have to check the continuity at x=−2,−1at x=−2LHL=limh→0(−2−h)2+3(−2−h)+2(−2−h)+2=limh→04+h2+4h−6−3h+2−h=limh→0h2+h−h=limh→0h+1−1=−1RHL=limh→0(−2+h)2+3(−2+h)+2(−2+h)+2=limh→04+h2−4h−6+3h+2h=limh→0h2−hh=limh→0h−11=−1⇒LHL=RHL=f(−2)∴ It is continuous at x=−2Now, check for x=−1LHL=h→0lim(−1−h)2+3(−1−h)+2(−1−h)+2=limh→012+h2+2h−3−3h+21−h=limh→0h2−h1−h=limh→02h−1−1=1RHL=limh→0(−1+h)2+3(−1+h)+2(−1+h)+2=limh→01+h2−2h−3+3h+21+h=limh→0h2+h1+h=1⇒LHL=RHL=f(−1)∴It is not continuous at x=−1The required function is continuous in R−{−1}.