How do you determine whether the function f(x)=xx2−5 is concave up or concave down and its intervals?

Step 1We have to find 2nd derivative,f′(x)=x2−5−x⋅2x(x2−5)2=−x2−5(x2−5)2fx)=−2x(x2−5)2−(−x2−5)2(x2−5)⋅2x(x2−5)4fx)=−2x(x2−5)−(−x2−5)2⋅2x(x2−5)3fx)=−2x3+10x+4x3+20x(x2−5)3Step 2fx)=2x3+30x(x2−5)3=2x(x2+15)(x2−5)3fx)=0⇔2x=0(x2+15≠0∀x∈R)fx)=0⇔x=0(x2−5)3&gt;0∀x∈(−∞,−5)∪(5,+∞)(x2−5)3&lt;0∀x∈(−5,5)fx)&gt;0∀x∈(−5,0)∪(5,+∞) function is concave upfx)&lt;0∀x∈(−∞,−5)∪(0,5) function is concave down

Key to "How do you determine whether the function f(x)=xx2−5..."

alahovom7ahc

Answered question

2022-04-15

How do you determine whether the function

f (x) = \frac{x}{x^{2} - 5}

is concave up or concave down and its intervals?

Answer & Explanation

Wrastirtyzp9w

Beginner2022-04-16Added 10 answers

Step 1
We have to find 2nd derivative,

f^{'} (x) = \frac{x^{2} - 5 - x \cdot 2 x}{{(x^{2} - 5)}^{2}} = \frac{- x^{2} - 5}{{(x^{2} - 5)}^{2}}

f x) = \frac{- 2 x {(x^{2} - 5)}^{2} - (- x^{2} - 5) 2 (x^{2} - 5) \cdot 2 x}{{(x^{2} - 5)}^{4}}

f x) = \frac{- 2 x (x^{2} - 5) - (- x^{2} - 5) 2 \cdot 2 x}{{(x^{2} - 5)}^{3}}

f x) = \frac{- 2 x^{3} + 10 x + 4 x^{3} + 20 x}{{(x^{2} - 5)}^{3}}

Step 2

f x) = \frac{2 x^{3} + 30 x}{{(x^{2} - 5)}^{3}} = \frac{2 x (x^{2} + 15)}{{(x^{2} - 5)}^{3}}

f x) = 0 \Leftrightarrow 2 x = 0 (x^{2} + 15 \neq 0 \forall x \in R)

f x) = 0 \Leftrightarrow x = 0

{(x^{2} - 5)}^{3} > 0 \forall x \in (- \infty, - \sqrt{5}) \cup (\sqrt{5}, + \infty)

{(x^{2} - 5)}^{3} < 0 \forall x \in (- \sqrt{5}, \sqrt{5})

f x) > 0 \forall x \in (- \sqrt{5}, 0) \cup (\sqrt{5}, + \infty)

function is concave up

f x) < 0 \forall x \in (- \infty, - \sqrt{5}) \cup (0, \sqrt{5})

function is concave down

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