Finding monotone interval and extremum of y = 2x Λ 3-3x Λ 2-12x + 14 by tabulation method

Finding monotone interval and extremum of y = 2x Λ 3-3x Λ 2-12x + 14 by tabulation method

We can derive the function
y'=6x^2-6x-12
=6(x-2)(x+1)
Then we can know the extremum of the function when x = 2 and x = - 1
x x0
Y increasing maximum decreasing minimum increasing
Then we can substitute it,
The function has a maximum of 21 when x = - 1 and a minimum of - 6 when x = 2

How to find the derivative of function y = 1 / (3x ^ 3)?

Answer: y '= - 1 / (x ^ 4)
y=1/(3x^3)=3x^-3
∴y'=3×(-1/3) x^(-3-1) =-x^-4=-1/(x^4)

Let f (x) = (1 / 2a) x ^ 2-lnx a be greater than zero
Finding monotone interval and extremum of F (x) when a is equal to 1

f(x)=(1/2a)x^2-lnx a=1
f(x)=(1/2)x^2-lnx f(x)′=x-(1/x)=(x^2-1)/x
If f (x) ′ > 0, X ＜ - 1 or X ＞ 1 ∵ x ＞ 0  x ＞ 1, then f (x) increases monotonically
If f (x) ′ ≤ 0, - 1 ≤ x ＜ 0 or 0 ＜ x ≤ 1 ‖ 0 ＜ x ≤ 1, then f (x) decreases monotonically
The extreme value is f (1) = 1 / 2

The function f (x) = LNX, G (x) = 1 / 2A (the square of x) + 2x is known;
Given the function f (x) = LNX, G (x) = 1 / 2A (the square of x) + 2x (1), if the function H (x) = f (x) - G (x) increases monotonically in the domain of definition, the value range of a is obtained
(2) Let's ask whether there is a real number a, so that the function K (x) = f (x) - G '(x) has two points on (1 / 2, positive infinity), and the tangents of these two points are perpendicular to each other. If there is a value of a, there is no reason

(1) H (x) = f (x) - G (x) is increasing, so h '(x) > = 0 h' (x) = 1 / X - (x / A + 2) > = 0 x > 0
1/x-x/a>=2 a