For what values of x is f(x)=−2x2x−1 concave or convex?

Step 1Given:&nbsp;f(x)=−2x2x−1f′(x)=(x−1)(−4x)−(−2x2)(1)(x−1)2=−2x2+4x(x−1)2fx)=(x−1)2(−4x+4)−(−2x2+4x)(2(x−1))(x−1)4fx)=−4(x−1)3Step 2The function will be concave if&nbsp;f(x)&lt;0∴−4(x−1)3&lt;01(x−1)3&gt;0x∈(1,∞)The function will be concave in&nbsp;(1,∞)The function will be convex in&nbsp;(−∞,1)

Answered: "For what values of x is f(x)=−2x2x−1 concave..."

High School
Calculus and Analysis
Calculus 1
Analyzing functions

Arianna Villegas

Answered question

2022-03-24

For what values of x is

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

concave or convex?

Answer & Explanation

slokningol09

Beginner2022-03-25Added 6 answers

Step 1
Given: $f (x) = \frac{- 2 x^{2}}{x - 1}$
$f^{'} (x) = \frac{(x - 1) (- 4 x) - (- 2 x^{2}) (1)}{{(x - 1)}^{2}}$
$= \frac{- 2 x^{2} + 4 x}{{(x - 1)}^{2}}$
$f x) = \frac{{(x - 1)}^{2} (- 4 x + 4) - (- 2 x^{2} + 4 x) (2 (x - 1))}{{(x - 1)}^{4}}$
$f x) = - \frac{4}{{(x - 1)}^{3}}$
Step 2
The function will be concave if $f (x) < 0$
$∴ - \frac{4}{{(x - 1)}^{3}} < 0$
$\frac{1}{{(x - 1)}^{3}} > 0$
$x \in (1, \infty)$
The function will be concave in $(1, \infty)$
The function will be convex in $(- \infty, 1)$

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