Question

PLEASE HELP NO ONE WILL HELP ME! I NEED PROS! TYSM! For f(x) = 3 x 5 − 5 x 3 + 3 . Determine the interval(s) on which f(x) is concave up and on which f(x) is concave down AND indicate where f(x) has inflection point(s). (If there are none, please tell me). Also, indicate where f(x) has local maximum and local minimum points. Also, can you please help me determine the interval(s) on which f(x) is increasing and on which f(x) is decreasing.

Answer 1

$\bf f(x)=3x^5-5x^3+3 \\\\\\ \cfrac{dy}{dx}=15x^4-15x^2\qquad \textit{critical points}\implies 0=15x^4-15x^2 \\\\\\ 0=x^4-x^2\implies 0=x^2(x^2-1)\implies x= \begin{cases} 0\\ -1\\ +1 \end{cases} \\\\\\ \cfrac{dy^2}{dx^2}= 60x^3-30x\qquad \textit{inflection points}\implies 0=60x^3-30x \\\\\\ 0=2x^3-x\implies 0=x(2x^2-1)\quad \begin{cases} 0=x\\ ----------\\ 0=2x^2-1\\ 1=2x^2\\ \frac{1}{2}=x^2\\ \pm\sqrt{\frac{1}{2}}=x\\\\ \pm\frac{\sqrt{2}}{2}=x \end{cases}$

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.