## Calculus (3rd Edition)

$f(x)$ is decreasing on $(-\infty,-0.6)$ and increasing on $(-0.6,\infty)$ . $x= -0.6$ is a local minimum
Given $$y=5 x^{2}+6 x-4$$ Since $$f'(x) = 10x+6$$ Then $f(x)$ has critical points when \begin{align*} f'(x)&=0\\ 10x+6&=0 \end{align*} Then $x= -0.6$ is a critical point. To find the interval where $f$ is increasing and decreasing, choose $x=0$, $x= -1$ \begin{align*} f'(0)&= 6>0\\ f'(-1)&= -4<0 \end{align*} Hence, $f(x)$ is decreasing on $(-\infty,-0.6)$ and increasing on $(-0.6,\infty)$. Since $f(x)$ changes from decreasing to increasing, then $x= -0.6$ is a local minimum.