Is #f(x)=3x^5-x^3+8x^2-x+3# concave or convex at #x=-4#?

Question

Ezekiel Alexander · Accepted Answer

Concave down

To find the concavity of a function #f(x)#, you will need to determine the second derivative #f''(x)#, and then evaluate at at the given point. If the result of #f''(a) > 0#, the function is concave up; if the result is negative, it it concave down. #f(x) = 3x^5 - x^3 + 8x^2 - x + 3# #f'(x) = 15x^4 - 3x^2 + 16x = 1# #f''(x) = 60x^3 - 6x + 16# #f''(-4) = 60(-4)^3 - 6(-4) + 16 = -3,800 < 0# Thus, #f(x)# is concave down at #x = -4# graph{3x^5 - x^3 + 8x^2 - x + 3 [-4.59, 2.72, -5000.82, 500.84]}

Christopher Allers · Answer

undefined To determine whether the function $ f(x) = 3x^5 - x^3 + 8x^2 - x + 3 $ is concave or convex at $ x = -4 $, we need to find the second derivative of the function and evaluate it at $ x = -4 $.

The second derivative of $ f(x) $ is denoted as $ f''(x) $, which represents the rate of change of the slope of the function.

First, find the first derivative $ f'(x) $ of $ f(x) $, then differentiate it again to find $ f''(x) $.

$ f'(x) = 15x^4 - 3x^2 + 16x - 1 $

$ f''(x) = 60x^3 - 6x + 16 $

Now, substitute $ x = -4 $ into the second derivative:

$ f''(-4) = 60(-4)^3 - 6(-4) + 16 $

$ f''(-4) = -960 - (-24) + 16 $

$ f''(-4) = -960 + 24 + 16 $

$ f''(-4) = -920 $

Since $ f''(-4) $ is negative, the function is concave at $ x = -4 $.