Find the Intervals of Concavity and Convexity for the Following Functions:

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Exercise 1

f (x) = 3x - x^3

 

Exercise 2

f (x) = x^4 - 2x^2 - 8

 

Exercise 3

f (x) = \frac{x^3}{(x - 1)^2}

 

Exercise 4

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

 

Exercise 5

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

 

Exercise 6

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

 

Exercise 7

f (x) = x + \sqrt{x}

 

Exercise 8

f (x) = e^ {-x^2}

 

Exercise 9

f (x) = e^{\frac{1}{x}}

 

Exercise 10

f (x) = (x - 1)e^ {-x}

 

Exercise 11

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

 

Exercise 12

f (x) = \frac{lnx}{x}

 

 

 

Solution of exercise 1

f (x) = 3x - x^3

f '' (x) = -6x

-6x = 0

x = 0

Convex: (-\infty, 0)

Concave: (0, \infty)

 

Solution of exercise 2

f(x) = x^4 - 2x^2 - 8

f '' (x) = 12x^2 - 4

12x^2 -4 = 0

x = \pm \frac{1}{\sqrt{3}} = \frac{\sqrt{3}} {3}

Concave: (-\infty, \frac{-\sqrt{3}} {3}) U (\frac{\sqrt{3}} {3}, \infty)

Convex: (-\frac{\sqrt{3}}{3}, \frac {\sqrt{3}}{3})

 

Solution of exercise 3

f (x) = \frac{x^3}{(x - 1)^2}

f ' (x) = \frac{x^3 - 3x^2}{(x - 1)^3}

f '' (x) = \frac{6x}{(x - 1)^4}

= \frac{6x}{(x - 1)^4} = 0

x = 0

Convex: (0, 1) U (1, \infty)

Concave: (-\infty, 0)

 

Solution of exercise 4

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

f ' (x) = \frac{2 (x^4 - 1)}{x^3}

f '' (x) = \frac{2 (x^4 + 3)}{x^4}

2 (x^4 + 3) = 0

x = \sqrt[4]{-3}

Convex: (-\infty, 0) U (0, \infty)

 

Solution of exercise 5

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

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

f '' (x) = \frac{8}{(2 - x)^3}

\frac{8}{(2 - x)^3} = 0

No solution

Convex: (-\infty, 2)

Concave: (2, \infty)

Solution of exercise 6

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

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

f '' (x) = \frac{2x^3 - 6x}{(1 + x^2)^3}

\frac{2x^3 - 6x}{(2 + x^2)^3} = 0

x = 0

x = \pm \sqrt{3}

Convex: (-\sqrt{3}, 0) U (\sqrt{3}, \infty)

Concave: (-\infty, -\sqrt{3}) U (0, \sqrt{3})

 

Solution of exercise 7

f (x) = x + \sqrt{x}

f ' (x) = 1 + \frac{1}{2\sqrt{x}}

f '' (x) = \frac{-1}{4x\sqrt{x}}

\frac{-1}{4x \sqrt{x}} = 0

No solution

Concave: (0, \infty)

 

Solution of exercise 8

f (x) = e^{-x^2}

f ' (x) = -2xe^{-x^2}

f '' (x) = -2 e^{-x^2} - 2x e ^ {-x^2} (-2x) = 2e^ {-x^2} (2x^2 - 1)

2e^{-x^2} (2x^2 - 1) = 0

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

Inflection points: (-\frac{\sqrt{2}}{2}, e^ {\frac{1}{2}})(\frac{\sqrt{2}}{2}, e ^ {\frac{1}{2}})

Convex: (-\infty, - \frac{\sqrt{2}} {2}) U (\frac{\sqrt{2}} {2}, \infty)

Concave: (-\frac{\sqrt{2}} {2}, \frac{\sqrt{2}} {2})

 

Solution of exercise 9

f (x) = e^ {\frac{1}{x}}

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

f '' (x) = \frac{1}{x^4} e ^ {\frac{1}{x}} (2x + 1)

\frac{1}{x^4} e^ {\frac{1}{x}} (2x + 1) = 0

e^{\frac{1}{x}} = 0         No solution

2x + 1 = 0        x = -\frac{1}{2}

Convex: (-\frac{1}{2}, 0) U (0, \infty)

Concave: (-\infty, - \frac{1}{2})

 

Solution of exercise 10

f (x) = (x - 1)e^{-x}

f ' (x) = e^{-x} (2 - x)

f '' (x) = e^{-x}(x - 3)

x - 3 = 0

x = 3

Convex: (3, \infty)

Concave: (-\infty, 3)

 

Solution of exercise 11

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

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

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

\frac{1}{2\pi} (x^2 - 1) e ^ {-\frac{1}{2}x^2} = 0

x^2 - 1 = 0

x = \pm 1

Convex: (-\infty, -1) U (1, \infty)

Concave: (-1, 1)

 

Solution of exercise 12

f (x) = \frac{lnx}{x}

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

f '' (x) = \frac{2 ln x - 3}{x^3}

\frac{2 ln x - 3 }{x^3} = 0

2lnx - 3 = 0

ln = \frac{3}{2}

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

Convex: (e ^ {\frac{3}{2}}, \infty)

Concave: (0, e^{\frac{3}{2}})

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.