Answer: \(y=\frac1{10}(3x+2)^{\frac53}+2x-1\).

Integrate the second derivative to find the gradient:

\(\frac{dy}{dx}=\int (3x+2)^{-\frac13}\,dx.\)

Using the reverse chain rule,

\(\frac{dy}{dx}=\frac12(3x+2)^{\frac23}+C.\)

The gradient is \(4\) when \(x=2\), so

\(4=\frac12(8)^{\frac23}+C.\)

Since \(8^{\frac23}=4\),

\(4=2+C,\)

and therefore \(C=2\). Thus

\(\frac{dy}{dx}=\frac12(3x+2)^{\frac23}+2.\)

Integrate again:

So

\(y=\frac1{10}(3x+2)^{\frac53}+2x+D.\)

The curve passes through \((2,6.2)\), so

\(6.2=\frac1{10}(8)^{\frac53}+4+D.\)

Since \(8^{\frac53}=32\),

\(6.2=3.2+4+D,\)

which gives \(D=-1\). Therefore

\(y=\frac1{10}(3x+2)^{\frac53}+2x-1.\)