Is #f(x)=x^3+2x^2-4x-12# concave or convex at #x=3#?

Question

Ezra Adams · Accepted Answer

At #f(3)# the function is convex.

A concave function is a function in which no line segment joining two points on its graph lies above the graph at any point. A convex function, on the other hand, is a function in which no line segment joining two points on the graph lies below the graph at any point. It means that, if #f(x)# is more than the average of #f(x+-lambda)# than the function is concave and if #f(x)# is less than the average of #f(x+-lambda)# than the function is convex. Hence to find the convexity or concavity of #f(x)=x^3+2x^2-4x-12# at #x=3#, let us evaluate #f(x)# at #x=2.5, 3 and 3.5#. #f(2.5)=(5/2)^3+2(5/2)^2-4(5/2)-12=125/8+50/4-10+12=(125+100-80-96)/8=49/8# #f(3)=(3)^3+2(3)^2-4(3)-12=27+18-12-12=21# #f(3.5)=(7/2)^3+2(7/2)^2-4(7/2)-12=343/8+98/4-14-12=(343+196-112-96)/8=331/8# The average of #f(2.5)# and #f(3.5)# is #(49/8+331/8)/2=380/(2xx8)=95/4=23 3/4# As, this is more than #f(3)#, at #f(3)# the function is convex. graph{x^3+2x^2-4x-12 [-5, 5, -20, 30]}

Axel Alvarez · Answer

undefined To determine the concavity of $ f(x) = x^3 + 2x^2 - 4x - 12 $ at $ x = 3 $, we need to find the second derivative of $ f(x) $ and evaluate it at $ x = 3 $.

First, find the first derivative of $ f(x) $:
$$ f'(x) = 3x^2 + 4x - 4 $$

Now, find the second derivative of $ f(x) $:
$$ f''(x) = 6x + 4 $$

Evaluate $ f''(3) $:
$$ f''(3) = 6(3) + 4 = 18 + 4 = 22 $$

Since $ f''(3) > 0 $, the second derivative is positive at $ x = 3 $, indicating that the graph of $ f(x) $ is concave up at $ x = 3 $.