Question

Let g(x) = {(ax^2-b) if x<2, (bx-a) if x>2}

Find a relationship between a and b (that is, solve for a in terms of b or vice versa) So that g(x) is continuous for all of x.

Answer 1

Since, g(x) is continuous for all of x.

Therefore, $(a.2^2-b)=b.2-a$ [LHL = RHL]

$4a-b=2b-a\Rightarrow 5a=3b \Rightarrow a=\frac{3}{5}b$