For what values of x is #f(x)=(-2x^2)/(x-1)# concave or convex?

Question

Cameron Alexander · Accepted Answer

The function will be concave in #(1, \infty)#

The function will be convex in #(- \infty, 1)# Given function: #f(x)=\frac{-2x^2}{x-1}# #f'(x)=\frac{(x-1)(-4x)-(-2x^2)(1)}{(x-1)^2}# #=\frac{-2x^2+4x}{(x-1)^2}# #f''(x)=\frac{(x-1)^2(-4x+4)-(-2x^2+4x)(2(x-1))}{(x-1)^4}# #f''(x)=-4/(x-1)^3# The function will be concave iff #f''(x)<0# # herefore -4/(x-1)^3< 0# #1/(x-1)^3> 0# #x\in (1, \infty)# The function will be concave in #(1, \infty)# The function will be convex in #(- \infty, 1)#

Paisley Johnson · Answer

undefined To determine the concavity or convexity of $ f(x) = \frac{-2x^2}{x-1} $, we first find its second derivative, $ f''(x) $. Then, we analyze the sign of $ f''(x) $ to determine concavity or convexity.

First, we find the first derivative:
$$ f'(x) = \frac{d}{dx} \left( \frac{-2x^2}{x-1} \right) = \frac{-2x^2 \cdot (x-1) - (-2x^2) \cdot 1}{(x-1)^2} = \frac{-2x^2 + 2x^2}{(x-1)^2} = \frac{0}{(x-1)^2} = 0 $$

Since the first derivative equals zero, we need to apply the quotient rule to find the second derivative:
$$ f''(x) = \frac{d}{dx} \left( \frac{-2x^2}{x-1} \right) = \frac{-2(x-1)(2x^2) - (-2x^2)(1)}{(x-1)^2} = \frac{-4x^2 + 4x^2}{(x-1)^2} = 0 $$

Since the second derivative equals zero, the concavity or convexity of $ f(x) $ is undetermined at this point. We need to examine the points of inflection, which occur when $ f''(x) = 0 $ or is undefined.

Setting $ f''(x) = 0 $, we have:
$$ -4x^2 + 4x^2 = 0 $$
$$ 0 = 0 $$

This equation is satisfied for all real values of $ x $. Therefore, there are no points of inflection for $ f(x) $.

Thus, the function $ f(x) = \frac{-2x^2}{x-1} $ is neither concave nor convex for any values of $ x $.