Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 10

y=(-3x^5+40x^3+135x)/270
differentiating
y'=1/270(-15x^4+120x^2+135)
differentiating again,
y''=1/270(-60x^3+240x)
y''=-60/270(x^3-4x)
y''=-60/270x(x+2)(x-2)
In order to determine the concavity , determine when y''=0,
So y''=0 for x=0 , x=-2 and x=2
Now test for concavity in the intervals (-oo ,-2) , (-2,0) , (0,2) and (2,oo )
Now let us plug in the test values x=-3 , -1 , 1 and 3
y''(-3) =-60/270(-3)(-1)(-5)=10/3
y''(-1)=-60/270(-1)(1)(-3)=-2/3
y''(1)=-60/270(1)(3)(-1)=2/3
y''(3)=-60/270(3)(5)(1)=-10/3
Since y''(-3) and y''(1) are positive ,so the graph is concave upward in the interval (-oo ,-2) and ((0,2)
y''(-1) and y''(3) are negative , so the graph is concave downward in the interval (-2.0) and (2,oo )