Examine if equation f(x)=0 have unique root in (1/e,e) if f(x)=x+lnx?

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You should check the signs of values of function at `x = 1/e` and `x = e` , such that:

`f(1/e) = 1/e + ln(1/e) => f(1/e) = 1/e + ln1 - lne`

`f(1/e) = 1/e + 0 - 1 => f(1/e) = 1/e - 1 < 0`

`f(e) = e + ln e = e + 1 > 0`

Since `f(1/e) < 0` and `f(e) > 0` , hence, there exists `x_0 in (1/e,e)` , such that `f(x_0) = 0` .

You need to prove that equation `f(x) = 0` has one solution in `(1/e,e)` , hence, you need to prove that the function f(x) strictly increases over `(1/e,e)` , thus, you need to prove that `f'(x) > 0` over `(1/e,e)` .

You need to evaluate `f'(x)` such that:

`f'(x) = (x + ln x)' => f'(x) = 1 + 1/x`

You need to evaluate `f'(x)` at `x = 1/e` and `x = e` , such that:

`f'(1/e) = 1 + 1/(1/e) => f'(1/e) = 1 + e > 0`

`f'(e) = 1 + 1/e > 0`

You should notice that `f'(x) > 0` for `x in (1/e,e)` , hence, the function strictly increases over `(1/e,e)` .

**Since the function increases over `(1/e,e)` , yields that the equation `f(x) = 0` has an unique solution over `(1/e,e)` .**

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