Get the solution to "How do you find concavity when f(x)=x73+x43?"

Saul Cochran

Answered question

2022-04-15

How do you find concavity when

f (x) = x^{\frac{7}{3}} + x^{\frac{4}{3}}

shanna87mn

Beginner2022-04-16Added 19 answers

Step 1
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) = \frac{7}{3} x^{\frac{4}{3}} + \frac{4}{3} x^{\frac{1}{3}}

f x) = \frac{28}{9} x^{\frac{1}{3}} + \frac{4}{9} x^{- \frac{2}{3}}

f x) = \frac{28 x + 4}{9 x^{\frac{2}{3}}}

Step 2
Notice how the

f x) = 0

when

x = - \frac{1}{7}

. This means that when

x = - \frac{1}{7}

there is a point of inflection on f(x).

f x) < 0

when

x < - \frac{1}{7}

, so f(x) is concave down whenever

x < - \frac{1}{7}

.
Similarly, f(x) is concave up whenever

x ≻ \frac{1}{7}

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