Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 365: 115

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Let's first find the value of x at which the graphs meet. At the point of intersection, $$e^x = e^{-x}$$ Or, $$e^{2x}=1$$ Hence $$x=0$$ at the point of intersection. At x = 0, slopes of the two graphs are $$m_1=\frac{de^x}{dx}|_{x=0}=e^0 =1$$ and $$m_2=\frac{de^{-x}}{dx}|_{x=0}=-e^{-0} =-1$$ Clearly, $$m_1\times m_2=-1$$ and hence the graphs intersect at right angles