CURVE SKETCHING

Question

. 𝑦 = 4𝑥^3 − 3𝑥^4

ANSWER: 𝐶𝑃: 𝑥 = 0 & 𝑥 = 1 , IP : 𝑥 = 0 ,2\3, Increasing on (−∞, 0) & (0,1) , Decreasing on (1, ∞), CU on(0,2\3), CD on (−∞, 0) & (2\3, ∞), Maximum point (1,1)

I WANT THE STEPS

Answer 1

This is the most difficult of your 3 posts, so I will do this one, you do the others in the same way.

y = 4x^3 - 3x^4
y' = dy/dx = 12x^2 - 12x^3
= 0 for max or min of y
12x^2 - 12x^3 = 0
12x^2(1 - x) = 0
x = 0 or x = 1
if x = 0 , y = 0
if x = 1, y = 4-3 = 1

y'' = 24x - 36x^2 = 0 at points of inflection
12x(2 - 3x) = 0
x = 0 or x = 2/3
when x = 0, y = 0
when x = 2/3, y = 4(2/3)^3 - 3(2/3)^4 = 16/27
So you have 2 points of inflection: (0,0), and (2/3 , 16/27)

increasing function: when y'>0
12x^2 - 12x^3 > 0
x^2 - x^3 > 0
x^2(1-x)>0
Since x^2 is always positive,
1-x > 0
x < 1
decreasing function: y' < 0
same argument, 1-x < 0, so x > 1

investigate if (0,0) is a max/min
Since (0,0) is also a point of inflection, it is neither a max nor a min

investigate (1,1)
at (1,1), y'' = 24(1) - 36(1) < 0, so by the 2nd derivative test,
(1,1) is a max.

When y''> 0, the function is concave up
when y'' < 0, it is concave down
24x - 36x^2 > 0
x - x^2 > 0
x(1-x)> 0 , <---- means either both positive or both negative.
case1:
x > 0 and 1-x > 0 -----> 0 < x < 1
case2:
x < 0 and 1-x < 0
x < 0 or x > 1

you can see all those properties showing up by using
www.desmos.com/calculator
and typing in your original equation