Question

Help me please??? I don’t know how to get started!

Answer 1

A differentiable function $f(x)$ is increasing on an open interval $(a,b)$ if $f'(x)>0$ for all $a$, and decreasing if $f'(x)$. For this problem, you then need to compute the derivative:

$f(x)=x^2\ln x\implies f'(x)=2x\ln x+x=(2\ln x+1)x$

then solve for $f'(x)=0$:

$(2\ln x+1)x=0\implies x=0\text{ or }x=e^{-1/2}$

We can ignore $x=0$ because $x^2\ln x$ is defined only for $x>0$. So we have two intervals to consider, $(0,e^{-1/2})$ and $(e^{-1/2},\infty)$. All we need to do is pick any value from either interval and check the sign of the derivative $f'(x)$. Since $e^{-1/2}\approx0.606$, from the first interval we can take $x=\dfrac12$, and from the second we can pick $x=1$.

$f'\left(\dfrac12\right)\approx-0.193$

$f'(1)=1>0$

The above indicates that $f(x)$ is decreasing on the first interval $(0,e^{-1/2})$, and increasing on the second interval $(e^{-1/2},\infty)$.

For part (b), we use the info from above as part of the first derivative test for extrema. We have one critical point at $x=e^{-1/2}$, and we know how $f(x)$ behaves to either side of this point; $f(x)$ decreases to left of it, and increases to the right. This pattern is indicative of a minimum occurring at $x=e^{-1/2}$, and we find that $f(x)$ has the (local) minimum value of $f(e^{-1/2})=-\dfrac1{2e}\approx-0.184$.