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Question Number 211637 by CrispyXYZ last updated on 15/Sep/24

$$f\left(x\right)=\sin x-{\mathrm{e}}^x+1.$$$$\mathrm{Prove}\mathrm{that}f\left(x\right)\mathrm{has}\mathrm{only}2\mathrm{zeros}\mathrm{in}-\pi ⩽x⩽0.$$

Answered by mehdee1342 last updated on 15/Sep/24

$$f\left(0\right)=0$$$$f(-\pi )=1-e^{-\pi }0\mathrm{\&}f(-\frac{\pi }{2})=-e^{-\frac{\pi }{2}}<0$$$$\Rightarrow \mathrm{\exists }-\pi <a<-\frac{\pi }{2};f\left(a\right)=0$$$$$$

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