1one
draw graph
7 Answers
11Well ,I think it has got just 1 root...
f(x)=e^{x-1}+x-2 \\ \ f(0)=\frac{1}{e}-2<0 \\ \ f'(x)=e^{x-1}+1>0 \\ \ x>0
Thus function crosses the X-axis once bet 1 and 2...
66As Soumik has shown, the derivative of the function f(x) (as defined in #6) is positive for all real x, so it is strictly increasing for all real x. Hence, it can cut the x axis exactly once. And obviously the root is x=1 as can be seen by direct substitution.
66Apart from it, one can also use Rolle's theorem. If there is, apart from x=1, any other root, say α, then consider f(x)=ex-1+x-2 in the interval [1, α]. We see that in this interval f(x) is continuous and in the interval (1, α) f(x) is differentiable. Further, f(1)=f(α)=0. Hence, by Rolle's theorem, there should exist at least one point c in the interval (1, α) where f'(c)=0. But that's a contradiction since f'(x)>0.