Is #f(x)=-2x^5+7x^2+8x-13# concave or convex at #x=-4#?

Question

Isabella Ackerman · Accepted Answer

Since #f''(-4) > 0#, #f(x)# is convex at #x = -4#.

Let's compute the second derivative: #f(x) color(white)(ii) = -2x^5 + 7x^2 + 8x - 13# #f'(x) color(white)(i) = -10 x^4 + 14 x + 8# #f''(x) = -40 x^3 + 14# Now, let's evaluate the second derivative at #x = -4# and check if #f''(-4)# is negative or positive: #f''(-4) = -40 * (-4)^3 + 14# Even without computing this value, you can see that #(-4)^3# is negative and #-40# is also negative. A negative value multiplied with a positive value is positive. #14# is positive as well. So we can conclude that #f''(-4) > 0# This means that #f(x)# is convex at #x = -4#.

Samantha Adkison · Answer

undefined To determine if the function $ f(x) = -2x^5 + 7x^2 + 8x - 13 $ is concave or convex at $ x = -4 $, we need to evaluate the second derivative of the function at that point. If the second derivative is positive, the function is convex at that point. If the second derivative is negative, the function is concave at that point.

The second derivative of $ f(x) $ is $ f''(x) = -40x^3 + 14 $.

Evaluate $ f''(-4) $:

$ f''(-4) = -40(-4)^3 + 14 $
$ f''(-4) = -40( -64) + 14 $
$ f''(-4) = 2560 + 14 $
$ f''(-4) = 2574 $

Since the second derivative $ f''(-4) = 2574 $ is positive, the function $ f(x) = -2x^5 + 7x^2 + 8x - 13 $ is convex at $ x = -4 $.