For what values of x is f(x)=4x2−1 concave or convex?

Check the solution to "For what values of x is f(x)=4x2−1 concave..."

Kiara Haas

2022-03-26

For what values of x is

f (x) = \frac{4}{x^{2} - 1}

concave or convex?

Cassius Villarreal

Beginner2022-03-27Added 11 answers

Step 1
The function is

f (x) = \frac{4}{x^{2} - 1}

The domain of

f (x)

x \in (- \infty, - 1) \cup (- 1, 1) \cup (1, + \infty)

Calulate the first derivative with the quotient rule

{(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}

u = 4, \Rightarrow, u^{'} = 0

v = x^{2} - 1, \Rightarrow, v^{'} = 2 x

Therefore,

f^{'} (x) = \frac{0 \cdot (x^{2} - 1) - 4 \cdot 2 x}{{(x^{2} - 1)}^{2}} = - \frac{8 x}{{(x^{2} - 1)}^{2}}

f^{'} (x) = 0, \Rightarrow, x = 0

There is a critical point at

(0, - 4)

Calulate the second derivative with the quotient rule

u = - 8 x, \Rightarrow, u^{'} = - 8

v = {(x^{2} - 1)}^{2}, \Rightarrow, v^{'} = 4 x (x^{2} - 1)

f^{″} (x) = \frac{- 8 {(x^{2} - 1)}^{2} + 32 x^{2} (x^{2} - 1)}{{(x^{2} - 1)}^{4}}

= \frac{- 8 x^{2} + 8 + 32 x^{2}}{{(x^{2} - 1)}^{3}}

= \frac{24 x^{2} + 8}{{(x^{2} - 1)}^{3}}

Therefore,

f^{″} (x) \neq 0, \forall x \in domain

Build a variation chart to determine the concavities

\begin{matrix} Interval & (- \infty, - 1) & (- 1, 1) & (1, + \infty) \\ Sign f^{″} (x) & + & - & + \\ f (x) & \cup & \cap & \cup \end{matrix}

Finally,

f (x)

is convex for

x \in (- \infty, - 1) \cup (1, + \infty)

f (x)

is concave for

x \in (- 1, 1)

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