# the function f(x) = (1+x)loge(1+x) show the curve y=f(x) has a turning point at (1-e/e) , (-1/e)

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You need to perform the first derivative test to find the turning points of the real valued function `f(x) = (1+x)ln(1+x)` such that:

`f'(x) = 0` .

You need to find the derivative of the function using quotient rule, such that:

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

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

Reducing duplicate factors yields:

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

You need to solve for x the equation f'(x) = 0 such that:

`ln(1 + x) + 1 = 0 => ln(1 + x) = -1 => 1 + x = e^(-1) => x = 1/e - 1`

You need to check what is the sign of function to the right of the value `x = (1 - e)/e` , hence, you need to evaluate f'(0) such that:

`f'(0) = ln1 + 1 = 1 > 0`

You need to evaluate `f((1-e)/e)` such that:

`f((1-e)/e) = (1 + (1-e)/e)(ln(1+(1-e)/e))`

`f((1-e)/e) = ((e+1-e)/e)(ln((e+1-e)/e))`

Reducing duplicate factors yields:

`f((1-e)/e) = (1/e)(ln(1/e)) => f((1-e)/e) = (1/e)ln(e^(-1)) => f((1-e)/e) = (1/e)*(-1)ln e => f((1-e)/e) = -1/e`

**Hence, evaluating the turning point of the given function yields that the function has a minimal turning point at `((1 - e)/e,-1/e).` **