Answer:
b) local minimum at [tex]x=\frac{1}{e}[/tex]
c) Graph of the function is convex
Step-by-step explanation:
We are told that [tex]f'(\frac{1}{e})=0[/tex], which implies there is a turning/stationary point at [tex]x=\frac{1}{e}[/tex].
Substituting [tex]x=\frac{1}{e}[/tex] into [tex]f''(x)[/tex] will tell us if the turning point is a minimum or a maximum:
[tex]f''(\frac{1}{e})=\frac{1}{\frac{1}{e} } =e>0 \implies \textsf{local minimum}[/tex]
Therefore, statement a) is false and statement b) is true.
If the function has a minimum turning point, then this implies that the curve is convex. Therefore, statement c) is true.
Extremum = local min and max points.
We have already established that there is a local minimum at [tex]x=\frac{1}{e}[/tex], therefore statement d) is false.
I know this is not needed for this question, but here are the workings to detemine the equation of the function (I've also attached a graph). This supports the answers above.
[tex]\textsf{if} \ f''(x)=\frac{1}{x}\\\\\implies f'(x)=\int f''(x) \ dx \\\\\implies f'(x)=ln|x|+C\\[/tex]
[tex]\textsf{if} \ f'(\frac{1}{e})=0\\\\\implies ln|\frac{1}{e}|+C=0\\\\\implies -1+C=0\\\\\implies C=1\\\\[/tex]
[tex]\implies f'(x)=ln|x|+1\\[/tex]
[tex]\textsf{if} \ f'(x)=ln|x|+1\\\\\implies f(x)=\int f'(x) \ dx\\\\\implies f(x)=xln(x)-x+x+C\\\\\implies f(x)=xln(x)+C[/tex]
