determine the open intervals on which the graph of is concave downward or concave upward. f(x)=x^2+2/x^2-4

Question

how do i solve this i am blank :( help please

Answer 1

Take the second derivative of f(x).  This means you will have to take the derivative of the first derivative and set it equal to zero.

 

f'(x) =

 

(2x(x² - 4) - 2x(x² + 2)) / (x² - 4)² =

 

(-8x - 4x) / (x² - 4)² =

 

-12x / (x² - 4)²

 

 

Now take the derivative of this and set it equal to zero.

 

(-12(x² - 4)² + 12x * 2(x² - 4) * 2x) / (x² - 4)⁴ = 0

 

(-12(x² - 4)² + 48x²(x² - 4)) / (x² - 4)⁴ = 0

 

(x² - 4)(-12(x² - 4) + 48x²) / (x² - 4)⁴ = 0

 

(36x² + 48) / (x² - 4)³ = 0

 

Since the numerator has no real solutions, there are no inflection points.  As proof of this, the function is not continuous.  Therefore, you should not have any inflections points.  Though, you will have concavity.

 

Your vertical asymptotes are x=-2 and x=2.  Finally, use test points to see where the 2nd derivative is positive and negative.

 

Evaluate the 2nd derivative at x=-3 , x=0 , and at x=3.  We can only use the vertical asymptotes as our reference points to test.

 

Concave up if 2nd derivative is positive.

Concave down if 2nd derivative is negative.

 

 

f''(-3) = positive value

f''(0) = negative value

f''(3) = positive value

 

Concave up in the interval (-∞, -2)∪(2, ∞)

Concave down in the interval (-2, 2)