Is #f(x)=x^5-4x^4+16x-4# concave or convex at #x=-2#?

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Answer 1 · 2017-03-18T14:44:06

The answer is #f(x)# is concave at #x=-2#

Let's calculate the first and second derivatives

#f(x)=x^5-4x^4+16x-4#

#f'(x)=5x^4-16x^3+16#

#f''(x)=20x^3-48x^2=4x^2(5x-12)#

Now, we calculate, #f''(-2)#

#f''(-2)=20*(-2)^3-48*(-2)^2#

#=-160-192=-352#

As, #f''(-2)<0#, we conclude that #f(-2)# is concave #nn#

Normally, we calculate the concavity over an interval for example

#]-oo, 0[#