Question

Find the value of k that will make the function continuous everywhere.
f(x)=3x+k x≤3
f(x)=kx^2-6 x>3

Answer 1

shouldn't it be (15/8) ?

Answer 2

the only discontinuity the function might have is at x=3, so for the limt to exist and the function to be continuous: \[\lim_{x \rightarrow 3} 3x+k = \lim_{x \rightarrow 3} kx^2 - 6\]\[9+k = 9 k - 6\]\[8k=15\]\[k=\frac{15}8\]