1st and second derivative test:
Find from the 1st derivative where the interval is increasing/decreasing. And find also the critical points.

Find from the 2nd derivative any points of inflection of the graph of the function and determine where the graph is concave upward and concave downward.

Support your answer by plotting in the same window the graph of the function and the inflectional tangents.

$f(x) = x^4 - 2x^3$

Really appreciating any help

1st and second derivative test:
Find from the 1st derivative where the interval is increasing/decreasing. And find also the critical points.

Find from the 2nd derivative any points of inflection of the graph of the function and determine where the graph is concave upward and concave downward.

Support your answer by plotting in the same window the graph of the function and the inflectional tangents.

$f(x) = x^4 - 2x^3$

Really appreciating any help
Hello,

calculate first the 1rst and 2nd derivative:

1) $f(x)=x^4-2x^3$

2) $f'(x)=4x^3-6x^2=x^2(4x-6)$

3) $f''(x)=12x^2-12x=12x(x-1)$

The conditions of critical points are: f'(x) = 0 and f''(x) ≠ 0

$x^2(4x-6)=0 \Longleftrightarrow x = 0 \ \vee \ x = \frac{3}{2}$

f(0) = 0 and f''(0) = 0 You've got a point of inflection with a horizontal gradient. Such a point is called in German (literally translated) a "terrace point".

f(3/2) = -27/16 and f''(3/2) = 9 > 0 that means the function has a minimum. From this point the function is continuously increasing.

The condition for the existence of points of inflection is $f''(x)=0$
From 3) you get x = 0 or x = 1

The 2 points are: $I_1(0, 0)\ \text{or}\ I_2(1, -1)$

The slope in I_1 is f'(0) = 0 therefore the tangent is the x-axis.

The slope in I_2 is f'(1) = -2. Use the point-slope-formula to get the equation of the tangent:

$y=-2x+1$

The graph is concave upward if f''(x) > 0 and the graph is concave downward if f''(x) < 0:

$12x(x-1) < 0 \Longrightarrow 0 < x < 1$

I've marked the concave downward part of the graph in red.

EB