Is #f(x)=(x-9)^3-x+15# concave or convex at #x=-3#?

Question

Christopher Agresta · Accepted Answer

#f(x)# is concave at #x=-3#

note: concave up = convex, concave down = concave

First we must find the intervals on which the function is concave up and concave down. We do this by finding the second derivative and setting it equal to zero to find the x values

#f(x) = (x-9)^3 - x + 15#

#d/dx = 3(x-9)^2 - 1#

#d^2/dx^2 = 6(x-9)#

#0 = 6x - 54#

#x = 9#

Now we test x values in the second derivative on either side of this number for positive and negative intervals. positive intervals correspond to concave up and negative intervals correspond to concave down

when x < 9: negative (concave down) when x > 9: positive (concave up)

So with the given x value of #x=-3#, we see that because #-3# lies on the left of 9 on the intervals, therefore #f(x)# is concave down at #x=-3#

Elijah Abram · Answer

undefined To determine the concavity of $ f(x) = (x-9)^3 - x + 15 $ at $ x = -3 $, we need to examine the sign of the second derivative at that point.

First, find the second derivative of $ f(x) $:

$$ f'(x) = 3(x-9)^2 - 1 $$
$$ f''(x) = 6(x-9) $$

Now, evaluate the second derivative at $ x = -3 $:

$$ f''(-3) = 6(-3-9) = 6(-12) = -72 $$

Since the second derivative $ f''(-3) = -72 $ is negative, the function is concave downwards at $ x = -3 $.