The figure above shows the graph of the function \(g\) and the line tangent to the graph of \(g\) at \(x=-1\). Let \(h\) be the function given by \(h(x)=e^x\cdot g(x)\). What is the value of \(h'(-1)\) ?

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$$ h(x)=e^x\cdot g(x) $$ $$ h'(x)=e^xg'(x)+g(x)e^x $$ $$ h'(x)=e^x(g'(x)+g(x)) $$

Use the graph to find \(g'(x)\) and \(g(x)\).

The derivative at \(g'(-1)\) is equal to the slope of the tangent line.

$$ m =\frac{-3-3}{0-(-1)} $$ $$ m = -\frac{6}{1} $$ $$ g'(-1) = -6 $$ $$ h'(-1)=e^{-1}(g'(-1)+g(-1)) $$ $$ h'(-1)=\frac{1}{e}(-6+3) $$ $$ = -\frac{3}{e} $$