Answer
(a) $f$ is decreasing on the interval $(0, \frac{1}{\sqrt{e}})$
$f$ is increasing on the interval $(\frac{1}{\sqrt{e}},\infty)$
(b) The local minimum is $f(\frac{1}{\sqrt{e}}) = -\frac{1}{2e}$
(c) The graph is concave down on this interval: $(0, \frac{1}{e^{3/2}})$
The graph is concave up on this interval: $(\frac{1}{e^{3/2}}~, \infty)$
The point of inflection is $(\frac{1}{e^{3/2}}~,~-\frac{3}{2e^3})$
Work Step by Step
(a) $f(x) = x^2~ln~x$
Note that this function is defined on the interval $(0, \infty)$
We can find the points where $f'(x) = 0$:
$f'(x) = 2x~ln~x+(x^2)(\frac{1}{x})$
$f'(x) = 2x~ln~x+x = 0$
$x~(2~ln~x+1) = 0$
$x=0$ or $ln~x = -\frac{1}{2}$
$x = 0, e^{-1/2}$
$x = 0, \frac{1}{\sqrt{e}}$
When $0 \lt x \lt \frac{1}{\sqrt{e}}$ then $f'(x) \lt 0$
$f$ is decreasing on the interval $(0, \frac{1}{\sqrt{e}})$
When $\frac{1}{\sqrt{e}} \lt x$ then $f'(x) \gt 0$
$f$ is increasing on the interval $(\frac{1}{\sqrt{e}},\infty)$
(b) $f(\frac{1}{\sqrt{e}}) = (\frac{1}{\sqrt{e}})^2~ln(\frac{1}{\sqrt{e}})$
$f(\frac{1}{\sqrt{e}}) = (\frac{1}{e})~ln(e^{-1/2})$
$f(\frac{1}{\sqrt{e}}) = (\frac{1}{e})~(-\frac{1}{2})$
$f(\frac{1}{\sqrt{e}}) = -\frac{1}{2e}$
The local minimum is $f(\frac{1}{\sqrt{e}}) = -\frac{1}{2e}$
(c) We can find the points where $f''(x) = 0$:
$f''(x) = 2~ln~x+(2x)(\frac{1}{x})+1$
$f''(x) = 2~ln~x+2+1$
$f''(x) = 2~ln~x+3 = 0$
$ln~x = -\frac{3}{2}$
$x = e^{-3/2}$
$x = \frac{1}{e^{3/2}}$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on this interval: $(0, \frac{1}{e^{3/2}})$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(\frac{1}{e^{3/2}}~, \infty)$
$f(\frac{1}{e^{3/2}}) = (\frac{1}{e^{3/2}})^2~ln(\frac{1}{e^{3/2}})$
$f(\frac{1}{e^{3/2}}) = (\frac{1}{e^3})~ln(e^{-3/2})$
$f(\frac{1}{e^{3/2}}) = (\frac{1}{e^3})~(-\frac{3}{2})$
$f(\frac{1}{e^{3/2}}) = -\frac{3}{2e^3}$
The point of inflection is $(\frac{1}{e^{3/2}}~,~-\frac{3}{2e^3})$
