Answer: \(\displaystyle f(x)=\frac89(3x+4)^{3/2}+2x-8\).

Answer: \(\displaystyle f(x)=\frac89(3x+4)^{3/2}+2x-8\).

We are given

\(f''(x)=6(3x+4)^{-1/2}.\)

Integrate to find \(f'(x)\):

\(f'(x)=4(3x+4)^{1/2}+C.\)

Use \(f'(4)=18\):

\(18=4(3(4)+4)^{1/2}+C.\)

Since \(3(4)+4=16\),

\(18=4\sqrt{16}+C=16+C.\)

Thus

\(C=2.\)

So

\(f'(x)=4(3x+4)^{1/2}+2.\)

Integrate again:

\(f(x)=\frac89(3x+4)^{3/2}+2x+D.\)

Use \(f(4)=\frac{512}{9}\):

\(\frac{512}{9}=\frac89(16)^{3/2}+2(4)+D.\)

Now \(16^{3/2}=64\), so

\(\frac{512}{9}=\frac89(64)+8+D.\)

Since \(\frac89(64)=\frac{512}{9}\),

\(D=-8.\)

Therefore

\(f(x)=\frac89(3x+4)^{3/2}+2x-8.\)