[tex]\bf f(x)=3x^5-5x^3+3
\\\\\\
\cfrac{dy}{dx}=15x^4-15x^2\qquad \textit{critical points}\implies 0=15x^4-15x^2
\\\\\\
0=x^4-x^2\implies 0=x^2(x^2-1)\implies x=
\begin{cases}
0\\
-1\\
+1
\end{cases}
\\\\\\
\cfrac{dy^2}{dx^2}= 60x^3-30x\qquad \textit{inflection points}\implies 0=60x^3-30x
\\\\\\
0=2x^3-x\implies 0=x(2x^2-1)\quad
\begin{cases}
0=x\\
----------\\
0=2x^2-1\\
1=2x^2\\
\frac{1}{2}=x^2\\
\pm\sqrt{\frac{1}{2}}=x\\\\
\pm\frac{\sqrt{2}}{2}=x
\end{cases}[/tex]
now, if you do a first-derivative test on those critical points, check the regions next to them, for example I checked x = 0.5 and x = -0.5, and both gave -2.8125, the value doesn't matter for the test, what matters is the sign, is negative, meaning on that region, the graph has a negative slope and thus is going downwards.
and then I checked x = 2 and x = -2, and both gave 180, which is positive, meaning the original graph is going up there, slope is increasing once you go passed the 1 or -1.
now, checking the inflection points when doing the second-derivative test, x = -1 gives -30, negative, concave down
x = -0.5 gives 7.5, positive, concave up
x = 0.5 gives -7.5, negative, concave down
x = 1 gives 30, positive, concave up
check the picture below, the arrows show the direction the slope going, and therefore how the original function is moving.
notice, before -1 is going up, reaches -1, then it goes down, meaning that's a peak, or maximum.
before 1 is going down, reaches 1 it dives, then goes back up, that's a minimum.
