How do you find concavity when #f(x)= x^(7/3) + x^(4/3)#?

Question

Emilia Aguilar · Accepted Answer

To find concavity, use the second derivative of the function.

If #f''(a)<0#, the function is concave down (also called concave) when #x=a#.

If #f''(a)>0#, the function is concave up (also called convex) when #x=a#.

If #f''(a)=0#, the point when #x=a# is when the concavity of a function shifts called the point of inflection.

With your function:

#f'(x)=7/3x^(4/3)+4/3x^(1/3)#

#f''(x)=28/9x^(1/3)+4/9x^(-2/3)#

#f''(x)=(28x+4)/(9x^(2/3))#

graph{(28x+4)/(9x^(2/3)) [-20.28, 20.27, -10.14, 10.14]}

This is a graph of #f''(x)#.

Notice how the #f''(x)=0# when #x=-1/7#. This means that when #x=-1/7# there is a point of inflection on #f(x)#.

#f''(x)<0# when #x<-1/7#, so #f(x)# is concave down whenever #x<-1/7#.

Similarly, #f(x)# is concave up whenever #x> -1/7#.

Charles Anctil · Answer

undefined To find the concavity of the function f(x) = x^(7/3) + x^(4/3), you need to determine the second derivative and then analyze its sign. 1. Find the first derivative of f(x) with respect to x. f'(x) = (7/3)x^(4/3) + (4/3)x^(1/3) 2. Find the second derivative by differentiating f'(x). f''(x) = (28/9)x^(1/3) + (4/9)x^(-2/3) 3. Analyze the sign of the second derivative to determine concavity. - If f''(x) > 0 for all x in the domain, the function is concave up. - If f''(x) < 0 for all x in the domain, the function is concave down. - If f''(x) changes sign at a point, there's a point of inflection. So, to determine concavity: - If x > 0, f''(x) > 0, so the function is concave up. - If x < 0, f''(x) < 0, so the function is concave down. - There's a point of inflection at x = 0 where the concavity changes.