Is #f(x)=-x^5-x^4-2x^3+x-7# concave or convex at #x=-2#?

Question

Elijah Abram · Accepted Answer

convex at x = -2

To test if a function is concave / convex at f(a), require to find the value of f''(a). • If f''(a) > 0 then f(x) is convex at x = a • If f''(a) < 0 then f(x) is concave at x = a hence f(x)# = -x^5 - x^4 - 2x^3 + x - 7# f'(x)# = -5x^4 - 4x^3 - 6x^2 + 1# and f''(x)# = -20x^3 - 12x^2 - 12x# #rArr f''(-2) = -20(-2)^3 - 12(-2)^2 - 12(-2)# = 160 - 48 + 24 = 136 since f''(-2) > 0 then f(x) is convex at x = -2 graph{(-x^5-x^4-2x^3+x-7) [-16.02, 16.02, -8.01, 8.01]}

Aurora Alvarado · Answer

undefined To determine whether the function $ f(x) = -x^5 - x^4 - 2x^3 + x - 7 $ is concave or convex at $ x = -2 $, we need to evaluate the second derivative of the function at that point.

First, find the first derivative of $ f(x) $:
$$ f'(x) = -5x^4 - 4x^3 - 6x^2 + 1 $$

Now, find the second derivative of $ f(x) $:
$$ f''(x) = -20x^3 - 12x^2 - 12x $$

Evaluate $ f''(-2) $:
$$ f''(-2) = -20(-2)^3 - 12(-2)^2 - 12(-2) = -160 - 48 + 24 = -184 $$

Since $ f''(-2) = -184 $ is negative, the function is concave at $ x = -2 $.