On what intervals the following equation is concave up, concave

Bronson Olson

Bronson Olson

Answered question

2022-04-24

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))?

Answer & Explanation

ophelialee4xn

ophelialee4xn

Beginner2022-04-25Added 14 answers

Step 1
To 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)>0, then f is concave up in a neighborhood of x0;
- if fx0)<0, then f is concave down in a neighborhood of x0;
- if fx0)=0 and the sign of f'' on a sufficiently small right-neighborhood of x0 is opposite to the sign of f'' 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 2
The second derivative is f(x)=7x6[8ln(x)+1]+x78x=x6[56ln(x)+15]
Let's study the positivity of f"(x):
- x6>0x0
- 56ln(x)+15>0ln(x)>1556x>e1556
So, considering that the domain is R+, we get that
- if 0<x<e1556 then fx)<0 is concave down;
- if x>e1556 then fx)>0 and f is concave up;
- if x=e1556 then fx)=0. Considering that on the left of this point f'' is negative and on the right it is positive, we conclude that x=e1556 is a (falling) inflection point

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