Answer: \(\displaystyle f(x)=\frac1{24}(5x+2)^{8/5}+3x-20\).
Differentiate first. Stationary points occur where the derivative is zero, while tangent and normal problems use the gradient at the given point.
We are given
\(f''(x)=(5x+2)^{-2/5}.\)
Integrating once,
\(f'(x)=\frac13(5x+2)^{3/5}+c.\)
Use \(f'(6)=\frac{17}{3}\). Since \(5(6)+2=32\),
\(\frac{17}{3}=\frac13(32)^{3/5}+c.\)
Now \(32^{3/5}=8\), so
\(\frac{17}{3}=\frac83+c.\)
Hence
\(c=3.\)
Therefore
\(f'(x)=\frac13(5x+2)^{3/5}+3.\)
Integrating again,
\(f(x)=\frac1{24}(5x+2)^{8/5}+3x+d.\)
Use \(f(6)=\frac{26}{3}\). Again \(5(6)+2=32\), and \(32^{8/5}=256\), so
\(\frac{26}{3}=\frac1{24}(256)+18+d.\)
Thus
\(\frac{26}{3}=\frac{32}{3}+18+d.\)
Since \(18=\frac{54}{3}\),
\(\frac{26}{3}=\frac{86}{3}+d.\)
So
\(d=-20.\)
Therefore
\(f(x)=\frac1{24}(5x+2)^{8/5}+3x-20.\)
