Explained: "Is f(x)=ex−x2x−1 concave or convex at x=0?"

Rubi Riggs

Answered question

2022-04-16

f (x) = e^{x} - \frac{x^{2}}{x - 1}

concave or convex at

x = 0

Jameson Jensen

Beginner2022-04-17Added 16 answers

Step 1
The derivative of a quotient is

(\frac{u}{v}) = \frac{u^{'} v - u v^{'}}{v^{2}}

The function is

f (x) = e^{x} - \frac{x^{2}}{x - 1}

The first derivative is

f^{'} (x) = e^{x} - \frac{2 x (x - 1) - x^{2} \cdot 1}{{(x - 1)}^{2}}

= e^{x} - \frac{x^{2} - 2 x}{{(x - 1)}^{2}}

Step 2
The second derivative is

f x) = e^{x} - \frac{(2 x - 2) {(x - 1)}^{2} - 2 (x^{2} - 2 x) (x - 1)}{{(x - 1)}^{2}}

= e^{x} - \frac{(2 x - 2) (x - 1) - (2 x^{2} - 4 x)}{{(x - 1)}^{3}}

= e^{x} - {\frac{2 x^{2} - 4 x + 2 - 2 x^{2} + 4 x}{x - 1}}^{3}}

= e^{x} - \frac{2}{{(x - 1)}^{3}}

Step 3
When

x = 0

f 0) = e^{0} - \frac{2}{{(- 1)}^{3}} = 1 - 2 = 3

f 0) > 0

, the function is convex at

x = 0

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