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

Question

William Abdalla · Accepted Answer

Study the sign of the 2nd derivative.

For #x<1# the function is concave.
For #x>1# the function is convex.

You need to study curvature by finding the 2nd derivative. #f(x)=-2x/(x-1)# The 1st derivative: #f'(x)=-2((x)'(x-1)-x(x-1)')/(x-1)^2# #f'(x)=-2(1*(x-1)-x*1)/(x-1)^2# #f'(x)=-2(x-1-x)/(x-1)^2# #f'(x)=2*1/(x-1)^2# The 2nd derivative: #f''(x)=(2*(x-1)^-2)'# #f''(x)=2((x-1)^-2)'# #f''(x)=2*(-2)(x-1)^-3# #f''(x)=-4/(x-1)^3# Now the sign of #f''(x)# must be studied. The denominator is positive when: #-(x-1)^3>0# #(x-1)^3<0# #(x-1)^3<0^3# #x-1<0# #x<1# For #x<1# the function is concave. For #x>1# the function is convex. Note: the point #x=1# was excluded because the function #f(x)# can not be defined for #x=1#, since the denumirator would become 0. Here is a graph so you can see with your eyes: graph{(-2x)/(x-1) [-14.08, 17.95, -7.36, 8.66]}

Paisley Johnson · Answer

undefined To determine the concavity of $ f(x) = \frac{-2x}{x-1} $, we need to find its second derivative and analyze its sign. First, find the first derivative of $ f(x) $: $$ f'(x) = \frac{d}{dx} \left( \frac{-2x}{x-1} \right) = \frac{-2(x-1) - (-2x)}{(x-1)^2} $$ $$ f'(x) = \frac{-2x + 2 + 2x}{(x-1)^2} = \frac{2}{(x-1)^2} $$ Now, find the second derivative: $$ f''(x) = \frac{d}{dx} \left( \frac{2}{(x-1)^2} \right) = \frac{-4(x-1)(-1) - 2(2(x-1))}{(x-1)^4} $$ $$ f''(x) = \frac{4(x-1) + 4(x-1)}{(x-1)^4} = \frac{8(x-1)}{(x-1)^4} $$ $$ f''(x) = \frac{8}{(x-1)^3} $$ Now, analyze the sign of $ f''(x) $: - $ f''(x) > 0 $ when $ (x-1)^3 > 0 $, which occurs when $ x > 1 $. - $ f''(x) < 0 $ when $ (x-1)^3 < 0 $, which occurs when $ x < 1 $. Therefore, $ f(x) = \frac{-2x}{x-1} $ is concave for $ x < 1 $ and convex for $ x > 1 $.