Resolved: "How do you describe the concavity of the..."

Joaquin Salas

Answered question

2022-04-16

How do you describe the concavity of the graph and find the points of inflection (if any) for

f (x) = x^{3} - 3 x + 2

shanna87mn

Beginner2022-04-17Added 19 answers

Step 1
The function has a minimum at

> x = 1

and the curve is concave upwards.
The function has a maximum at

> x = - 1

and the curve is concave downwards
Explanation:
Given:

y = x^{3} - 3 x + 2

\frac{d y}{d x} = 3 x^{2} - 3

\frac{d^{2} x}{{d x}^{2}} = 6 x

\frac{d y}{d x} = 0 \Rightarrow 3 x^{2} - 3 = 0

3 x^{2} = 3

x^{2} = \frac{3}{3} = 1

\sqrt{x^{2}} = \pm \sqrt{1}

x = 1

x = - 1

Step 2
At

x = 1

;

\frac{d^{2} x}{{d x}^{2}} = 6 (1) = 6 > 0

x = 1; \frac{d y}{d x} = 0; \frac{d^{2} x}{{d x}^{2}} > 0

Hence the function has a minimum at

x = 1

and the curve is concave upwards.
At

x = - 1

;

\frac{d^{2} x}{{d x}^{2}} = 6 (- 1) = - 6 < 0

x = - 1; \frac{d y}{d x} = 0; \frac{d^{2} x}{{d x}^{2}} < 0

Hence the function has a maximum at

x = - 1

and the curve is concave downwards .

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