If #f(x)=(x^2+6x+8)/(x+2)# if x<-2 and #kx^2#if #x>= -2#, what value of k would make the function continuous at x=-2?

1 Answer
Jul 18, 2016

Answer:

#k=1/2#.

Explanation:

For #x<-2, f(x)=(x^2+6x+8)/(x+2)=((x+4)(x+2))/(x+2)#, and, since,

#x!=-2#, we can cancel #(x+2)# to get,

#f(x)=((x+4)cancel(x+2))/cancel(x+2)=x+4, if, x<-2#

Now, as #xrarr -2-, x<-2, so, f(x)=x+4#

#:. lim_(xrarr-2-)f(x)=lim_(xrarr-2-)(x+4)=-2+4=2..........(1)#

Further, as #xrarr-2+, x>-2, so, f(x)=kx^2#

#:. lim_(#xrarr-2+) f(x)=lim_(#xrarr-2+)kx^2=4k...................(2)#

Also, #f(-2)=4k....................................(3)#

Because #f# is reqd. to be continuous at #x=-2#, by defn. of continuity, we must have,

#lim_(xrarr-2-)f(x)=f(-2)=lim_(xrarr-2+)f(x)#.

#(1),(2) and (3) rArr 2=4k rArr k=1/2#.

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