Question 1142315
<br>
f(x) = 3x^5-5x^3+3<br>
f'(x) = 15x^4-15x^2 = 15x^2(x^2-1) = 15(x^2)(x-1)(x+1)<br>
f''(x) = 60x^3-30x = 30x(x^2-1) = 30(x)(x-1)(x+1)<br>
The critical points for both first and second derivatives are -1, 0, and 1; the resulting intervals are (-infinity, -1), (-1,0), (0,1), and (1, infinity).<br>
An analysis, using the factored forms of the first and second derivatives and test points in each interval:<pre>
   interval               f'(x)                           f''(x)
 ---------------------------------------------------------------------------------
  (-infinity, -1)  (+)(+)(-)(-) = +  increasing    (+)(-)(-)(-) = -  concave down
  (-1,0)           (+)(+)(+)(-) = -  decreasing    (+)(-)(+)(-) = +  concave up
  (0,1 )           (+)(+)(+)(-) = -  decreasing    (+)(+)(+)(-) = -  concave down
  (1,infinity)     (+)(+)(+)(+) = -  increasing    (+)(+)(+)(+) = +  concave up</pre>
A graph confirming the preceding analysis:<br>
{{{graph(400,400,-2,2,-10,10,3x^5-5x^3+3)}}}<br>
NOTE: While most resources use test points in each interval to determine increasing/decreasing and concave up/concave down, there is an easier and faster way to make those determinations.<br>
Consider "walking" along the number line and seeing what happens to the signs of the factors of the derivatives each time you pass one of the critical points.<br>
First derivative test for increasing/decreasing:<br>
For large negative values of x, f'(x) = 15(x^2)(x-1)(x+1) is positive; the function is increasing.
When we pass x=-1, one factor in f'(x) changes sign, so the sign of f'(x) changes; the function is now decreasing.
When we pass x=0, TWO factors change sign, resulting in no change to the sign of f'(x); the function is still decreasing.
When we pass x=1,  one factor changes sign, so the sign of f'(x) changes; the function is now increasing.<br>
Summary: increasing on (-infinity,-1) and (1,infinity); decreasing on (-1,0) and (0,1).<br>
Second derivative test for concave up/concave down:<br>
For large negative values of x, f''(x) = 30(x)(x-1)(x+1) is negative; the function is concave down.
When we pass each of the three critical points, one factor in f''(x) changes sign, so the concavity changes at each critical point.<br>
Summary: concave down on (-infinity,-1) and (0,1); concave up on (-1,0) and (1,infinity).