To two decimal places, find the value of k that will make the function f (x) continuous everywhere. 3x+k

solution:

we are consider the following function,

f (x) = 3x+k, x/leq 3

=kx^{2}-6, x>3

/lim_{x/rightarrow 3^-} (3x+k) = 9+k

/lim_{x/rightarrow3^+}

(kx^{2}-6) = 9k-6

so the left and right limits are equal.

therefore, the function is continuous at x=3

so, the therom of the function is continous at x=3

9k-6=9+k

8k=15

k=15/8

=1.875

therefore, the value of k=1.875