What is the value of k+c if f(x)= {2kx^2-x,x>3 and x^3+cx,x<=3} is everywhere deffirentiable

Question

The function is a peacewise function where 2kx^2 has a domain of x>3 and x^3+cx has a domain of x<= 3

Answer 1

We need to establish continuity at x=3.

So 2kx²-x=x³+cx so 18k-3=27+3c, 18k-3c=30, 6k-c=10, so c=6k-10

The gradients must also be the same at x=3, so

4kx-1=3x²+c at x=3, 12k-1=27+c, 12k-c=28, so c=12k-28.

Therefore 12k-28=6k-10, 6k=18, k=3 and c=8.

Now we have continuity and differentiability.