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xe^-x + e^-x = 0
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Solve for x?
yes
\[\Large\bf\sf x\color{#DD4747 }{e^{-x}}+\color{#DD4747 }{e^{-x}}\quad=\quad 0\] Factor an e^-x out of each term: \[\Large\bf\sf \color{#DD4747 }{e^{-x}}(x+1)\quad=\quad 0\]
Does that first step make sense? Understand why there is a +1?
yes
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Then to solve for x, we'll use the Zero Factor Property. We'll set each factor equal to zero and solve for x separately in each case.\[\Large\bf\sf e^{-x}=0,\qquad\qquad\qquad (x+1)=0\]
In the first case, we have an exponential function. Exponential function never touches the x-axis. We have no solution for x there. How bout the other one? Subtract 1 from each side, yes? :O
x=-1
thank you!
Yesssss, good job \c:/
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