Question

What are all values of x for which the graph of y= 6x^2 + x/2 + 3 + 6/x is concave downward?

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Answer 1

Answer:

$(-\sqrt[3]{3} , 0)$ in interval notation

Explanation:

f(x) = 6x^2 + 0.5x + 3 + 6/x

f'(x) = 12x + 0.5 – 12/(x^2)

f''(x) = 12 +36/(x^3)

Find where the second derivative is zero or undefined

x = 0 since 36/0 is undefined

x = [tex]-\sqrt[3]{3}[/tex] since:

-12 = 36/(x^3)

x^3 = -3

x = $-\sqrt[3]{3}$

Now find the intervals where the second derivative by using a number line with the values zero and negative cube root of three

Using this method, we can only see that when x is between $-\sqrt[3]{3}$ and 0, the function is concave down.