Behavior of a function

Solution:

a.)    The function y = f(x) is increasing at point x₀ if f‘(x₀) > 0

b.)    The function y = f(x) is decreasing at point x₀ if f‘(x₀) < 0

c.)    The function y = f(x) has a stationary point at x₀ if f‘(x₀) = 0

d.)    The function y = f(x) is convex at x₀ if f‘‘(x₀) > 0

e.)    The function y = f(x) is concave at x₀ if f‘‘(x₀) < 0

f.)    The function y = f(x) has an inflection point at x₀ if f‘‘(x₀) = 0 ^ f‘‘(x₀) ≠ 0

g.)    The function y = f(x) has a local minimum at x₀ if f‘ (x₀) = 0 ^ f‘‘ (x₀) > 0

h.)    The function y = f(x) has a local maximum at x₀ if f‘ (x₀) = 0 ^ f‘‘ (x₀) < 0

Solution:  

Stationary points:
$y = x^3 - 5x^2 + 3x - 5$
$y^{\mathrm{\prime }}=3x^2-10x+3$
$3x^2 - 10x + 3 = 0$

$3(x - 3)\left(x - \frac{1}{3}\right) = 0$
$x_1=3$ or $x_2 = \frac{1}{3}$

Increasing function:

$3(x-3)(x-\frac{1}{3})>0$

Divide by 3:
$(x-3)(x-\frac{1}{3})>0$

Using the sign analysis of the quadratic expression:

So:
$x > 3$ or $x < \frac{1}{3}$

Decreasing function:

Inflection point:

$y'' = 6x - 10$

$6x - 10 = 0$

Convex function:

$6x-10>0$

Concave function:

$6x-10<0$

Graph 

Solution:

y = x³ -5x² +3x -5

y‘ = 3x² -10x +3

y‘(2) = 3·2² -10·2 +3

y‘(2) = -5 < 0

The function is decreasing.

y‘ = 3x² -10x +3

y‘‘ = 6x - 10

y‘‘(2) = 6·2 – 10

y‘‘(2) = 2 > 0

The function is convex.