Question

Please help for this question what is the function and thanks

Answer 1

Answer:

b) local minimum at $x=\frac{1}{e}$

c) Graph of the function is convex

Step-by-step explanation:

We are told that $f'(\frac{1}{e})=0$, which implies there is a turning/stationary point at $x=\frac{1}{e}$.

Substituting $x=\frac{1}{e}$ into $f''(x)$ will tell us if the turning point is a minimum or a maximum:

$f''(\frac{1}{e})=\frac{1}{\frac{1}{e} } =e>0 \implies \textsf{local minimum}$

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 $x=\frac{1}{e}$, 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.

$\textsf{if} \ f''(x)=\frac{1}{x}\\\\\implies f'(x)=\int f''(x) \ dx \\\\\implies f'(x)=ln|x|+C$

$\textsf{if} \ f'(\frac{1}{e})=0\\\\\implies ln|\frac{1}{e}|+C=0\\\\\implies -1+C=0\\\\\implies C=1$

$\implies f'(x)=ln|x|+1$

$\textsf{if} \ f'(x)=ln|x|+1\\\\\implies f(x)=\int f'(x) \ dx\\\\\implies f(x)=xln(x)-x+x+C\\\\\implies f(x)=xln(x)+C$