Given f(x)=x^4-27/x^2, for what values of x is the graph of concave downwards? Multiple Choice- 10pts

Use the second derivative test for concavity:

f'(x)=(4x^3(x^2)-x^4-27(2x))/x^4
f'(x)=(4x^5-2x^5+54x)/x^4
f'(x)=(2x^4+54)/x^3
f''(x)=(8x^3(x^3)-3x^2(2x^4+54))/x^6
f''(x)=(8x^6-6x^6-162x^2)/x^6
f''(x)=(2x^4-162)/x^4
When you set f"(x) equal to zero you get x=0,3,-3 so those are your points of interception. You need to test values for x in the intervals -inf f''(-4)= Positive value, so it is concave up.
f''(-1)= Negative value, so it is concave down.
f''(1)=Negative value, so it is concave down.
f''(4)= positive value, so it is concave up.

The answer would be D.

the an

f(x)=(x^4-27)(x^2)

need 2nd derivative

f'(x) = 4x^3/x2 +-(2/x^3)(x^4-27) = 4x -2x -54/x^3 = 2x-54/x^3

f''(x) = 2 - 162/x^4

F''(X) = 0 = 2 -162/X^4 162/X^4 = 2 81 = X^4 X = -3, X = 3 are the points of inflection

f''(-4) is positive, meaning the graph is concave up -inf,-3

f(4) is also positive concave upward 3,inf

so f is concave down -3