How do you find concavity when $f (x) = x^{\frac{7}{3}} + x^{\frac{4}{3}}$ $f(x)= x^(7/3) + x^(4/3)$ ?

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Answer 1 · 2015-12-06T20:23:27

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 $x> -1/7$ .

How do you find concavity when f(x)= x^(7/3) + x^(4/3)f(x)=x73+x43f(x)= x^(7/3) + x^(4/3)?