On what intervals the following equation is concave up, concave down and where it's inflection point is (x,y)f(x)=x8(ln⁡(x))?

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On what intervals the following equation is concave up, concave down and where it&#039;s inflection point is (x,y)f(x)=x8(ln⁡(x))?

ophelialee4xn · Accepted Answer

Step 1To analyze concavity and inflection points of a twice differentiable function f, we can study the positivity of the second derivative. In fact, if x0 is a point in the domain of f, then:- if fx0)&amp;gt;0, then f is concave up in a neighborhood of x0;- if fx0)&amp;lt;0, then f is concave down in a neighborhood of x0;- if fx0)=0 and the sign of f&#039;&#039; on a sufficiently small right-neighborhood of x0 is opposite to the sign of f&#039;&#039; on a sufficiently small left-neighborhood of x0, then x=x0 is called an inflection point of f.In the specific case of f(x)=x8ln⁡(x). we have a function whose domain has to be restricted to the positive reals R+.The first derivative is f′(x)=8x7ln⁡(x)+x813=x7[8ln⁡(x)+1]Step 2The second derivative is f″(x)=7x6[8ln⁡(x)+1]+x78x=x6[56ln⁡(x)+15]Let&#039;s study the positivity of f&quot;(x):- x6&amp;gt;0⇔x≠0- 56ln⁡(x)+15&amp;gt;0⇔ln⁡(x)&amp;gt;−1556⇔x&amp;gt;e1556So, considering that the domain is R+, we get that- if 0&amp;lt;x&amp;lt;e−1556 then fx)&amp;lt;0 is concave down;- if x&amp;gt;e−1556 then fx)&amp;gt;0 and f is concave up;- if x=e−1556 then fx)=0. Considering that on the left of this point f&#039;&#039; is negative and on the right it is positive, we conclude that x=e−1556 is a (falling) inflection point

On what intervals the following equation is concave up, concave

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