Can anyone find all values of x for which the graph of f(x) = (x + 2)(x - 5)^2 is concave upward?

Question

1/3 < x < 5 ... Is this correct?

Answer 1

f''(x) = 6x - 16, from your previous question, so x has an inflection point at 8/3. The graph of f is concave upward when f''(x) > 0, which is the case when x > 8/3.

The values you gave appear to be where the graph of f is decreasing, since f'(x) < 0 on that interval (this is off the top of my head, so this should be checked, but 5 and 1/3 are the roots of f'

Answer 2

When you find the 1st der of f(x) is 3x^2-16x+5 and let that be 0 (zero) and you will find that the station pts are x=1/3 and 5. However when you sub both pts into the 2nd of f(x) which is 6x-16, x=5 gives you the answer of the 2nd der > 0, it therefore means that the point is going to be a minimum, another word its going to be a concave upward, whereas when x=1/3 it will give you a concave downward (max)
Furthermore, concave upward happens when f"(x)>0 so x>8/3 in order to be concave upward. And if you know, the fn has a point of inflexion at x=8/3

Answer 3

f(x) = (x+2)(x-5)^2 = x^4 - 9x^2 - a hundred If that facilitates, i'm satisfied, it is fairly plenty so far as i'm able to circulate devoid of employing the "guess and examine" technique from the available solutions.