Answer: \(\frac{dy}{dx}=(2x+3)^{1/2}+5x-8\); \(\mathrm f(x)=\frac13(2x+3)^{3/2}+\frac52x^2-8x+2\).

(a) Since

\(\frac{d^2y}{dx^2}=(2x+3)^{-\frac12}+5,\)

integrate once to find the gradient:

\(\frac{dy}{dx}=(2x+3)^{\frac12}+5x+C.\)

The curve has gradient \(10\) when \(x=3\), so

\(10=(2(3)+3)^{\frac12}+5(3)+C.\)

Thus

\(10=3+15+C,\)

so \(C=-8\). Therefore

\(\frac{dy}{dx}=(2x+3)^{\frac12}+5x-8.\)

When \(x=11\),

\(\frac{dy}{dx}=(25)^{\frac12}+55-8=5+55-8=52.\)

(b) Integrate the expression for the gradient:

Since

we get

The curve passes through \(\left(3,\frac{19}{2}\right)\), so

This gives

\(\frac{19}{2}=9+\frac{45}{2}-24+D.\)

So

\(D=2.\)

Therefore