Question

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?

Answer 1

`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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