Answer: (a) \(A=-12\), \(B=2\); (b) approximate change \(-6.8h\).

Work through the problem step by step, using exact values where possible before giving any final numerical approximation.

(a) Let

\(u=\sqrt{3x^2-2}\).

Then

\(y=\frac{u}{x-4}\).

Differentiate \(u\):

\(\frac{du}{dx}=\frac{1}{2}(3x^2-2)^{-1/2}\cdot6x\).

So

\(\frac{du}{dx}=\frac{3x}{\sqrt{3x^2-2}}\).

Using the quotient rule,

\(\frac{dy}{dx}=\frac{(x-4)\frac{du}{dx}-u}{(x-4)^2}\).

Substitute \(u=\sqrt{3x^2-2}\):

\(\frac{dy}{dx}=\frac{(x-4)\frac{3x}{\sqrt{3x^2-2}}-\sqrt{3x^2-2}}{(x-4)^2}\).

Put the numerator over the common denominator \(\sqrt{3x^2-2}\):

\(\frac{dy}{dx}=\frac{3x(x-4)-(3x^2-2)}{(x-4)^2\sqrt{3x^2-2}}\).

Simplify the numerator:

\(3x(x-4)-(3x^2-2)=3x^2-12x-3x^2+2=-12x+2\).

Therefore

\(\frac{dy}{dx}=\frac{-12x+2}{(x-4)^2\sqrt{3x^2-2}}\).

So \(A=-12\) and \(B=2\).

(b) When \(x=3\),

\(\sqrt{3x^2-2}=\sqrt{27-2}=5\),

and

\((x-4)^2=(3-4)^2=1\).

Thus

\(\frac{dy}{dx}=\frac{-12(3)+2}{1\cdot5}=\frac{-34}{5}=-6.8\).

For a small increase \(h\) in \(x\), the approximate change in \(y\) is

\(\frac{dy}{dx}\,h\).

Therefore the approximate change in \(y\) is

\(-6.8h\).