Is #f(x)=x^4-4x^3+x-4# concave or convex at #x=-1#?

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Answer 1 · 2016-03-22T20:11:20

convex

To find that out, we need to get the second derivative first.

Getting the first derivative.

#[1]" "f'(x)=d/dx(x^4-4x^3+x-4)#

We can easily find this using power rule.

#[2]" "f'(x)=4x^3-12x^2+1#

Getting the second derivative.

#[1]" "f''(x)=d/dx(4x^3-12x^2+1)#

Use power rule again.

#[2]" "f''(x)=12x^2-24x#

Now that we know the second derivative, we will evaluate #f''(x)# at #x=-1# to check its concavity.

• If #f''(x)>0#, then it is concave up or convex
• If #f''(x)<0#, then it is concave down or concave

#[1]" "f''(-1)=12(-1)^2-24(-1)#

#[2]" "f''(-1)=12+24#

#[3]" "f''(-1)=36#

Since #f(x)# is 36 at #x=-1# and 36 is greater than 0, then #f(x)# is convex at #x=-1#.