Calculus (3rd Edition)

$f(x)$ is decreasing on $(-\infty,-3/4)$ and increasing on $( -3/4,0) \cup (0 ,\infty)$. $f(x)$ has a local minimum at $x= -3/4$.
Given $$y=x^{4}+x^{3}$$ Since $$f'(x) =4x^{3}+3x^2$$ Then $f(x)$ has critical points when \begin{align*} f'(x)&=0\\ 4x^{3}+3x^2 &=0\\ x^2(4x +3)&=0 \end{align*} Then $x= 0$ and $x=-3/4$ are critical points. To find the intervals where $f$ is increasing and decreasing, choose $x=-1,\ x= -0.5$ and $x= 1$ \begin{align*} f'( -1)&<0 \\ f'(-0.5)& >0\\ f'(1)&>0 \end{align*} Hence, $f(x)$ is decreasing on $(-\infty,-3/4)$ and increasing on $( -3/4,0) \cup (0 ,\infty)$. Hence, by using the first derivative test, $f(x)$ has a local minimum at $x= -3/4$. $f(0)$ is a point of inflection.