Let \(f\) and \(g\) be differentiable functions with the following properties:

$$ \begin{array}{c l} (\text{i}) & g(x) \gt 0 \text{ for all } x \\ (\text{ii}) & f(0) = 1 \end{array} $$

If \(h(x)=f(x)g(x) \text{ and } h'(x)=f(x)g'(x)\), then \(f(x) =\)

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In general, for \(h(x)=f(x)g(x)\), we would use the product rule to obtain:

$$ h'(x)=f(x)g'(x)+f'(x)g(x) $$

Notice that the question specifies that \(h'(x)\) is only equal to the first part of our derivative. For this to be true, the second part has to be equal to zero:

$$ \colorbox{yellow}{$h'(x)=f(x)g'(x)$}+\colorbox{chartreuse}{$f'(x)g(x)$} $$ $$ \colorbox{chartreuse}{$f'(x)g(x) = 0$} \text { if } h'(x)=f(x)g'(x) $$

The question specifies that \(g(x)\) is always positive, so \(f'(x)\) must equal 0 so that the entire expression is 0.

$$ \colorbox{chartreuse}{$f'(x)g(x) = 0$} $$ $$ f'(x)=0 \text{ if } g(x)>0 $$

If \(f'(x)=0\), the graph of \(f\) must be a horizontal line.

Since \(f(0)=1\), \(f(x)=1\) for all values \(x\).