For what values of x is #f(x)= -5x^3+x^2+4x-12 # concave or convex?

Question

For what values of x is #f(x)= -5x^3+x^2+4x-12  # concave or convex?

William Abdalla · Accepted Answer

#f(x)# is convex on #(-oo,1/15)#, concave on #(1/15,+oo)#, and has a point of inflection when #x=1/15#.

#f(x)# is convex when #f''(x)>0#. #f(x)# is concave when #f''(x)<0#. Find #f''(x)#: #f(x)=-5x^3+x^2+4x-12# #f'(x)=-15x^2+2x+4# #f''(x)=-30x+2# The concavity could change when #f''(x)=0#. This is a possible point of inflection. #f''(x)=0# #-30x+2=0# #x=1/15# Analyze the sign surrounding the point #x=1/15#. You can plug in test points to determine the sign. When #x<1/15#, #f''(x)>0#. When #x>1/15#, #f''(x)<0#. When #x=15#, #f''(x)=0#. Thus, #f(x)# is convex on #(-oo,1/15)#, concave on #(1/15,+oo)#, and has a point of inflection when #x=1/15#. Graph of #f(x)#: graph{-5x^3+x^2+4x-12 [-22.47, 28.85, -20.56, 5.1]} The concavity does seem to shift very close to #x=0#.

Elijah Abram · Answer

undefined To determine where the function $ f(x) = -5x^3 + x^2 + 4x - 12 $ is concave or convex, we need to find the second derivative of the function and then analyze its sign. The second derivative of $ f(x) $ is denoted as $ f''(x) $. We find it by taking the derivative of the first derivative of $ f(x) $. $ f'(x) = -15x^2 + 2x + 4 $ $ f''(x) = -30x + 2 $ For the function to be concave upwards (convex), $ f''(x) $ must be positive. For it to be concave downwards, $ f''(x) $ must be negative. $ f''(x) > 0 $ when $ -30x + 2 > 0 $ $ -30x + 2 = 0 $ $ x = \frac{1}{15} $ So, $ f''(x) > 0 $ when $ x < \frac{1}{15} $. Hence, the function is convex for $ x < \frac{1}{15} $. $ f''(x) < 0 $ when $ -30x + 2 < 0 $ $ -30x + 2 = 0 $ $ x = \frac{1}{15} $ So, $ f''(x) < 0 $ when $ x > \frac{1}{15} $. Hence, the function is concave for $ x > \frac{1}{15} $.