Is #f(x)=-x^3+2x^2-4x-2# concave or convex at #x=0#?

1 Answer
Jan 10, 2016

If #f(x)# is a function, then to find that the function is concave or convex at a certain point we first find the second derivative of #f(x)# and then plug in the value of the point in that. If the result is less than zero then #f(x)# is concave and if the result is greater than zero then #f(x)# is convex.

That is,

if #f''(0)>0#, the function is convex when #x=0#
if #f''(0)<0#, the function is concave when #x=0#

Here #f(x)=-x^3+2x^2-4x-2#

Let #f'(x)# be the first derivative

#implies f'(x)=-3x^2+4x-4#

Let #f''(x)# be the second derivative

#implies f''(x)=-6x+4#

Put #x=0# in the second derivative i.e #f''(x)=-6x+4#.

#implies f''(0)=-6*0+4=0+4=4#
#implies f''(0)=4#

Since the result is greater then #0# therefore the function is convex.