Given f(x) =(x)(e^(-x)), find x when f '''(x) - f ''(x)=0 I got solutions as x=0 & x= 3 and was wondering if that's the correct answer.
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Originally Posted by tenner Given f(x) =(x)(e^(-x)), find x when f '''(x) - f ''(x)=0 I got solutions as x=0 & x= 3 and was wondering if that's the correct answer. $\displaystyle f'''(x)=(3-x)\cdot e^{-x} ~\text { and }~ f''(x)=(x-2) \cdot e^{-x}$ Hence : $\displaystyle f'''(x)-f''(x)=0$ $\displaystyle (3-x)\cdot e^{-x}-(x-2) \cdot e^{-x}=0$ $\displaystyle (5-2x)\cdot e^{-x}=0$ $\displaystyle x=\frac{5}{2}$
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