# `y = (-3x^5 + 40x^3 + 135x)/(270)` Determine the open intervals on whcih the graph is concave upward or downward.

### Textbook Question

Chapter 3, 3.4 - Problem 10 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`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`