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

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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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differentiating again,




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




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`

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