For what values of x is f(x)=−2xx−1 concave or convex?

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star04iks7 · Accepted Answer

Study the sign of the 2nd derivative.For x&amp;lt;1 the function is concave.For x&amp;gt;1 the function is convex.Explanation:Step 1You need to study curvature by finding the 2nd derivative.f(x)=−2xx−1The 1st derivative:f′(x)=−2(x)′(x−1)−x(x−1)′(x−1)2f′(x)=−21⋅(x−1)−x⋅1(x−1)2f′(x)=−2x−1−x(x−1)2f′(x)=2⋅1(x−1)2Step 2The 2nd derivative:fx)=(2⋅(x−1)−2),fx)=2((x−1)−2),fx)=2⋅(−2)(x−1)−3fx)=−4(x−1)3Step 3Now the sign of f&quot;(x) must be studied. The denominator is positive when:−(x−1)3&amp;gt;0(x−1)3&amp;lt;0(x−1)3&amp;lt;03x−1&amp;lt;0x&amp;lt;1For x&amp;lt;1 the function is concave.For x&amp;gt;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.

For what values of x is \(\displaystyle{f{{\left({x}\right)}}}={\frac{{-{2}{x}}}{{{x}-{1}}}}\)

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