Aryanna Rowland

2022-03-16

What is the instantaneous rate of change of $f\left(x\right)=\left({x}^{2}-3x\right){e}^{x}$ at x=2?

orangepaperiz7

The instantaneous rate of change when x=2 can be found through computing
f'(2).
To find f'(2), first find f'(x)
Use the product rule:
${f}^{\prime }\left(x\right)={e}^{x}\frac{d}{dx}\left[{x}^{2}-3x\right]+\left({x}^{2}-3x\right)\frac{d}{dx}\left[{e}^{x}\right]$
${f}^{\prime }\left(x\right)={e}^{x}\left(2x-3\right)+{e}^{x}\left({x}^{2}-3x\right)$
${f}^{\prime }\left(x\right)={e}^{x}\left(2x+3+{x}^{2}-3x\right)$
${f}^{\prime }\left(x\right)={e}^{x}\left({x}^{2}-x+3\right)$
${f}^{\prime }\left(2\right)={e}^{2}\left({2}^{2}-2+3\right)$
${f}^{\prime }\left(2\right)=5{e}^{2}$
${f}^{\prime }\left(2\right)=5{e}^{2}$
${f}^{\prime }\left(2\right)\approx 36.945$

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