In: Math

If \(f(x)=e^{x} g(x),\) where \(g(0)=3\) and \(g(0)=4,\) find \(f(0)\)

since \(f(x)=e^{x} g(x)\)

Then \(f^{\prime}(x)=e^{x}\left\{\frac{d}{d x}(g(x))\right\}+\left\{\frac{d}{d x}\left(e^{x}\right)\right\} g(x)\) (by product rule)

$$ \begin{array}{c} =e^{x} g^{\prime}(x)+e^{x} g(x) \\ \text { So, } f(0)=e^{0} g^{\prime}(0)+e^{0} g(0)=1 * 4+1 * 3=7 \quad\left(\text { as } e^{0}=1\right) \end{array} $$

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