The tangent to the curve $y = e^{x}$ drawn at the point $(c, e^{c})$ intersects the line joining the points $(c - 1, e^{c-1})$ and $(c + 1, e^{c+1})$

$y=e^{x}, \quad P=(c,\,e^{c})$ $A=(c-1,\,e^{c-1}), \quad B=(c+1,\,e^{c+1})$ $\text{Tangent at } P \ \cap \ \text{line } AB = \,?$ $(1)\ x<c, \quad (2)\ x>c, \quad (3)\ \text{never}, \quad (4)\ \text{always}$ $1.\ \text{tangent at } P$ $2.\ \text{chord } AB$ $3.\ \text{solve } t(x)=L(x) \Rightarrow x<c \text{ or } x>c$ $m_{\text{tan}}=e^{c}$ $t(x)=e^{c}+e^{c}(x-c)=e^{c}(x-c+1)$ $M=\left(c,\ \dfrac{e^{c+1}+e^{c-1}}{2}\right), \quad m_{AB}=\dfrac{e^{c+1}-e^{c-1}}{2}$ $L(x)=\dfrac{e^{c+1}+e^{c-1}}{2}+m_{AB}\,(x-c)$ $t(x)=L(x) \Rightarrow e(u+1)=\dfrac{e^{2}+1}{2}+\dfrac{e^{2}-1}{2}\,u, \quad u=x-c$ $2e(u+1)=(e^{2}+1)+(e^{2}-1)u$ $u\,(e^{2}-2e-1)=-(e-1)^{2}$ $u=-\dfrac{(e-1)^{2}}{e^{2}-2e-1}$ $e^{2}-2e-1=(e-1)^{2}-2>0 \Rightarrow u<0 \Rightarrow x<c$ $c=0:\ t(x)=x+1, \quad L(x)\approx 1.543+1.175x$ $x+1=1.543+1.175x \Rightarrow x\approx -3.10<0=c \ \checkmark$ $\boxed{\text{left of } x=c \ \Rightarrow\ \text{option }(1)}$