Answer: \(\displaystyle f(x)=\frac{1}{10}(2-3x)^{5/3}-4x-1\).

Answer: \(\displaystyle f(x)=\frac{1}{10}(2-3x)^{5/3}-4x-1\).

Integrate the second derivative:

\(\displaystyle \frac{dy}{dx}=\int(2-3x)^{-1/3}\,dx\).

Let \(u=2-3x\). Then \(du=-3\,dx\), so

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

The gradient at \((-2,10.2)\) is \(-6\). When \(x=-2\),

\((2-3x)^{2/3}=8^{2/3}=4\).

So

\(-6=-\frac12(4)+C\),

which gives \(C=-4\).

Hence

\(\displaystyle \frac{dy}{dx}=-\frac12(2-3x)^{2/3}-4\).

Integrate again:

\(\displaystyle y=\frac{1}{10}(2-3x)^{5/3}-4x+D\).

Use the point \((-2,10.2)\). Since \((2-3(-2))^{5/3}=8^{5/3}=32\),

\(10.2=\frac{32}{10}+8+D\).

Thus \(D=-1\).

Therefore

\(\displaystyle f(x)=\frac{1}{10}(2-3x)^{5/3}-4x-1\).