Is #f(x)=x(x-2)(x+3)(x-1)# concave or convex at #x=1 #?

Question

Is #f(x)=x(x-2)(x+3)(x-1)# concave or convex at #x=1   #?

Emilia Aguilar · Accepted Answer

#f(x)# is concave DOWNWARDS (convex?) at #x=1#.

This problem may be easier if we determine the expanded (not factored) form of #f(x)# first. #f(x)=x(x-2)(x+3)(x-1)=x(x^2+x-6)(x-1)=x(x^3-7x+6)=x^4-7x^2+6x# #f^'(x)=4x^3-14x+6# and #f^('')(x)=12x^2-14# #f^('')(1)=12*1^2-14=-2# Which means that #f(x)# is concave DOWNWARDS (convex?) at #x=1# because #f^('')(1)<0# graph{x(x-2)(x+3)(x-1) [-4, 3, -26, 10]}

Isaiah Allen · Answer

undefined To determine the concavity or convexity of the function $ f(x) = x(x-2)(x+3)(x-1) $ at $ x = 1 $, we need to evaluate the second derivative of the function at $ x = 1 $. If the second derivative is positive at $ x = 1 $, then the function is concave up (convex) at that point. If the second derivative is negative at $ x = 1 $, then the function is concave down at that point.

The second derivative of $ f(x) $ with respect to $ x $ is given by:

$$ f''(x) = 12x^2 - 6x - 6 $$

Now, plug in $ x = 1 $:

$$ f''(1) = 12(1)^2 - 6(1) - 6 = 12 - 6 - 6 = 0 $$

Since the second derivative at $ x = 1 $ is zero, we cannot determine the concavity or convexity of the function at that point using the second derivative test. We need to employ other methods such as the first derivative test or analyzing the behavior of the function around $ x = 1 $.