The functions \(f\), \(g\) and \(h\) are related by \[f'(x) = g(x+1), \qquad g'(x)=h(x-1).\] It follows that \(f''(2x)\) equals
\(h(2x+1)\);
\(2h'(2x)\);
\(h(2x)\);
\(4h(2x)\).
Calculus meets Functions
Question
The functions \(f\), \(g\) and \(h\) are related by \[f'(x) = g(x+1), \qquad g'(x)=h(x-1).\] It follows that \(f''(2x)\) equals
\(h(2x+1)\);
\(2h'(2x)\);
\(h(2x)\);
\(4h(2x)\).